#set document(title: "3.1 Obtuse Angles", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.1#h(0.6em)Obtuse Angles The town of Avery lies 48 miles due east of Baker, and Clio is 34 miles from Baker, in the direction #math.equation(block: false, alt: "35 degrees")[$35^(∘)$] west of north. How far is it from Avery to Clio? We know how to solve right triangles using the trigonometric ratios. But the triangle formed by the three towns is not a right triangle, because it includes an obtuse angle of #math.equation(block: false, alt: "125 degrees")[$125^(∘)$] at #math.equation(block: false, alt: "B")[$B$], as shown in the figure. #figure(figph[triangle of three towns], alt: "triangle of three towns", caption: none) A triangle that is not a right triangle is called an #strong[oblique] triangle. In this chapter we learn how to solve oblique triangles using the laws of sines and cosines. But first we must be able to find the sine, cosine, and tangent ratios for obtuse angles. === Angles in Standard Position To extend our definition of the trigonometric ratios to obtuse angles, we use a Cartesian coordinate system. We put an angle #math.equation(block: false, alt: "θ")[$θ$] in #strong[standard position] as follows: - Place the vertex at the origin with the #strong[initial side] on the positive #math.equation(block: false, alt: "x")[$x$]-axis; - the #strong[terminal side] opens in the counter-clockwise direction. - We choose a point #math.equation(block: false, alt: "P")[$P$] on the terminal side of the angle, and form a right triangle by drawing a vertical line from #math.equation(block: false, alt: "P")[$P$] to the #math.equation(block: false, alt: "x")[$x$]-axis. The length of the side adjacent to #math.equation(block: false, alt: "θ")[$θ$] is the #math.equation(block: false, alt: "x")[$x$]-coordinate of point #math.equation(block: false, alt: "P")[$P$], and the length of the side opposite is the #math.equation(block: false, alt: "y")[$y$]-coordinate of #math.equation(block: false, alt: "P")[$P$]. The length of the hypotenuse is the distance from the origin to #math.equation(block: false, alt: "P")[$P$], which we call #math.equation(block: false, alt: "r")[$r$]. With this notation, our definitions of the trigonometric ratios are as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Coordinate Definitions of the Trigonometric Ratios] - #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r")[$cos ( θ ) = display(frac(x, r))$] - #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r")[$sin ( θ ) = display(frac(y, r))$] - #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x")[$tan ( θ ) = display(frac(y, x))$] #figure(figph[angle in standard position], alt: "angle in standard position", caption: none) ] It doesn't matter which point #math.equation(block: false, alt: "P")[$P$] on the terminal side we use to calculate the trig ratios. If we choose some other point, say #math.equation(block: false, alt: "P prime")[$P^(′)$], with coordinates #math.equation(block: false, alt: "open parenthesis x prime , y prime close parenthesis")[$( x^(′) , y^(′) )$], as shown at right, we will get the same values for the sine, cosine and tangent of #math.equation(block: false, alt: "θ")[$θ$]. The new triangle formed is similar to the first one, so the ratios of the sides of the new triangle are equal to the corresponding ratios in the first triangle. #figure(figph[similar triangles], alt: "similar triangles", caption: none) #examplebox("Example 1")[][ Find the values of #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis")[$cos ( θ ) , " " sin ( θ )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] if the point #math.equation(block: false, alt: "open parenthesis 12 , 5 close parenthesis")[$( 12 , 5 )$] is on the terminal side of #math.equation(block: false, alt: "θ")[$θ$]. #solutionbox[ For the point #math.equation(block: false, alt: "P open parenthesis 12 , 5 close parenthesis")[$P ( 12 , 5 )$], we have #math.equation(block: false, alt: "x equals 12")[$x = 12$] and #math.equation(block: false, alt: "y equals 5")[$y = 5$]. We use the distance formula to find #math.equation(block: false, alt: "r")[$r$]. #math.equation(block: true, alt: "r, equals the square root of open parenthesis 2 minus 0 close parenthesis squared plus open parenthesis 5 minus 0 close parenthesis squared; equals the square root of 25 plus 144 equals the square root of 169 equals 13")[$r & = sqrt(( 2 − 0 )^(2) + ( 5 − 0 )^(2)) \ & = sqrt(25 + 144) = sqrt(169) = 13$] The trig ratios are #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis, equals the fraction x over r equals the fraction 12 over 13; sin ⁡ open parenthesis θ close parenthesis, equals the fraction y over r equals the fraction 5 over 13; tan ⁡ open parenthesis θ close parenthesis, equals the fraction y over x equals the fraction 5 over 12")[$cos ( θ ) & = display(frac(x, r)) = display(frac(12, 13)) \ sin ( θ ) & = display(frac(y, r)) = display(frac(5, 13)) \ tan ( θ ) & = display(frac(y, x)) = display(frac(5, 12))$] #figure(figph[angle], alt: "angle", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the previous Example, we get the same results by using the triangle definitions of the trig ratios. We can create a right triangle by dropping a vertical line from #math.equation(block: false, alt: "P")[$P$] to the #math.equation(block: false, alt: "x")[$x$]-axis, as shown in the figure. The legs of the right triangle have lengths 12 and 5, and the hypotenuse has length 13. ] + Find the equation of the terminal side of the angle in the previous example. (Hint: The terminal side lies on a line that goes through the origin and the point #math.equation(block: false, alt: "open parenthesis 12 , 5 close parenthesis")[$( 12 , 5 )$].) + Show that the point #math.equation(block: false, alt: "P prime open parenthesis 24 , 10 close parenthesis")[$P^(′) ( 24 , 10 )$] also lies on the terminal side of the angle. + Compute the trig ratios for #math.equation(block: false, alt: "θ")[$θ$] using the point #math.equation(block: false, alt: "P prime")[$P^(′)$] instead of #math.equation(block: false, alt: "P")[$P$]. + #math.equation(block: false, alt: "y equals the fraction 5 over 12 x")[$y = display(frac(5, 12)) x$] + #math.equation(block: false, alt: "open parenthesis 24 , 10 close parenthesis")[$( 24 , 10 )$] satisfies #math.equation(block: false, alt: "y equals the fraction 5 over 12 x")[$y = display(frac(5, 12)) x$], that is, the equation #math.equation(block: false, alt: "10 equals the fraction 5 over 12 open parenthesis 24 close parenthesis")[$10 = display(frac(5, 12)) ( 24 )$] is true. + #math.equation(block: false, alt: "r squared equals 24 squared plus 10 squared equals 676")[$r^(2) = 24^(2) + 10^(2) = 676$], so #math.equation(block: false, alt: "r equals the square root of 676 equals 26.")[$r = sqrt(676) = 26.$] Then #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r equals the fraction 24 over 26 equals the fraction 12 over 13; sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r equals the fraction 10 over 26 equals the fraction 5 over 13; tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x equals the fraction 10 over 24 equals the fraction 5 over 12")[$cos ( θ ) = display(frac(x, r)) = display(frac(24, 26)) = display(frac(12, 13)) \ sin ( θ ) = display(frac(y, r)) = display(frac(10, 26)) = display(frac(5, 13)) \ tan ( θ ) = display(frac(y, x)) = display(frac(10, 24)) = display(frac(5, 12))$] #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Obtuse Angles] Recall that an #strong[obtuse angle] has measure between #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. Use the coordinate definitions of the trig ratios described above to complete the Activity. #figure(figph[grid], alt: "grid", caption: none) + + Draw an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position with the point #math.equation(block: false, alt: "P open parenthesis 6 , 4 close parenthesis")[$P ( 6 , 4 )$] on its terminal side. + Find #math.equation(block: false, alt: "r")[$r$], the distance from the origin to #math.equation(block: false, alt: "P")[$P$]. + Calculate #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis , cos ⁡ open parenthesis θ close parenthesis")[$sin ( θ ) , " " cos ( θ )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Give both exact answers and decimal approximations rounded to four places. + Use the inverse cosine key on your calculator to find #math.equation(block: false, alt: "θ")[$θ$]. Use your calculator to verify the values of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis , cos ⁡ open parenthesis θ close parenthesis")[$sin ( θ ) , " " cos ( θ )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] that you found in part (3). + + Draw another angle #math.equation(block: false, alt: "ϕ")[$ϕ$] in standard position with the point #math.equation(block: false, alt: "Q open parenthesis minus 6 , 4 close parenthesis")[$Q ( − 6 , 4 )$] on its terminal side. + Explain why #math.equation(block: false, alt: "ϕ")[$ϕ$] is the supplement of #math.equation(block: false, alt: "θ")[$θ$]. (Hint: Consider the right triangles formed by drawing vertical lines from #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$].) + Can you use the right triangle definitions (using opposite, adjacent and hypotenuse) to compute the sine and cosine of #math.equation(block: false, alt: "ϕ")[$ϕ$]? Why or why not? + Calculate #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis , cos ⁡ open parenthesis ϕ close parenthesis")[$sin ( ϕ ) , " " cos ( ϕ )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis ϕ close parenthesis")[$tan ( ϕ )$] using the coordinate definitions. How are the trig values of #math.equation(block: false, alt: "ϕ")[$ϕ$] related to the trig values of #math.equation(block: false, alt: "θ")[$θ$]? + + Explain why #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "ϕ")[$ϕ$] have the same sine but different cosines. + Use the inverse cosine key on your calculator to find #math.equation(block: false, alt: "ϕ")[$ϕ$]. Use your calculator to verify the values of #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis , cos ⁡ open parenthesis ϕ close parenthesis")[$sin ( ϕ ) , " " cos ( ϕ )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis ϕ close parenthesis")[$tan ( ϕ )$] that you found above. + Compute #math.equation(block: false, alt: "180 degrees minus ϕ")[$180^(∘) − ϕ$]. What answer should you expect to get? ] === Trigonometric Ratios for Obtuse Angles Our new definitions for the trig ratios work just as well for obtuse angles, even though #math.equation(block: false, alt: "θ")[$θ$] is not technically “inside” a triangle, because we use the coordinates of #math.equation(block: false, alt: "P")[$P$] instead of the sides of a triangle to compute the ratios. Notice first of all that because #math.equation(block: false, alt: "x")[$x$]-coordinates are negative in the second quadrant, the cosine and tangent ratios are both negative for obtuse angles. For example, in the figure below, the point #math.equation(block: false, alt: "open parenthesis minus 4 , 3 close parenthesis")[$( − 4 , 3 )$] lies on the terminal side of the angle #math.equation(block: false, alt: "θ")[$θ$]. We see that #math.equation(block: false, alt: "r equals the square root of open parenthesis minus 4 close parenthesis squared plus 3 squared equals 5")[$" " " " r = sqrt(( − 4 )^(2) + 3^(2)) = 5 " " " "$], so #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis, equals the fraction x over r equals the fraction minus 4 over 5; sin ⁡ open parenthesis θ close parenthesis, equals the fraction y over r equals the fraction 3 over 5; tan ⁡ open parenthesis θ close parenthesis, equals the fraction y over x equals the fraction 3 over minus 4 equals the fraction minus 3 over 4")[$cos ( θ ) & = display(frac(x, r)) = display(frac(− 4, 5)) \ sin ( θ ) & = display(frac(y, r)) = display(frac(3, 5)) \ tan ( θ ) & = display(frac(y, x)) = display(frac(3, − 4)) = display(frac(− 3, 4))$] #figure(figph[obtuse angle], alt: "obtuse angle", caption: none) #examplebox("Example 2")[][ Find the values of #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] if #math.equation(block: false, alt: "θ")[$θ$] is an obtuse angle with #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 3")[$sin ( θ ) = display(frac(1, 3))$]. #solutionbox[ Because #math.equation(block: false, alt: "θ")[$θ$] is obtuse, the terminal side of the angle lies in the second quadrant, as shown in the figure below. Because #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 3")[$sin ( θ ) = display(frac(1, 3))$], we know that #math.equation(block: false, alt: "the fraction y over r equals the fraction 1 over 3")[$display(frac(y, r)) = display(frac(1, 3))$], so we can choose a point #math.equation(block: false, alt: "P")[$P$] with #math.equation(block: false, alt: "y equals 1")[$y = 1$] and #math.equation(block: false, alt: "r equals 3")[$r = 3$]. To find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] we need to know the value of #math.equation(block: false, alt: "x")[$x$]. From the Pythagorean Theorem, #math.equation(block: true, alt: "x squared plus 1 squared, equals 3 squared; x squared, equals 3 squared minus 1 squared equals 8; x, equals minus the square root of 8")[$x^(2) + 1^(2) & = 3^(2) \ x^(2) & = 3^(2) − 1^(2) = 8 \ x & = − sqrt(8)$] #figure(figph[obtuse angle], alt: "obtuse angle", caption: none) Remember that #math.equation(block: false, alt: "x")[$x$] is negative in the second quadrant! Thus #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r equals the fraction minus the square root of 8 over 3 and tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x equals the fraction minus 1 over the square root of 8")[$cos ( θ ) = display(frac(x, r)) = display(frac(− sqrt(8), 3)) " " " " " " " " " " "and" " " " " " " " " " " tan ( θ ) = display(frac(y, x)) = display(frac(− 1, sqrt(8)))$] ] ] + Sketch an obtuse angle #math.equation(block: false, alt: "θ")[$θ$] whose cosine is #math.equation(block: false, alt: "the fraction minus 8 over 17")[$display(frac(− 8, 17))$]. + Find the sine and the tangent of #math.equation(block: false, alt: "θ")[$θ$]. + By the Pythagorean theorem, #math.equation(block: true, alt: "y squared equals 17 squared minus open parenthesis minus 8 close parenthesis squared equals 15 squared")[$y^(2) = 17^(2) − ( − 8 )^(2) = 15^(2)$], so #math.equation(block: false, alt: "y equals 15")[$y = 15$]. #figure(figph[obtuse angle], alt: "obtuse angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 15 over 17 , tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 15 over 8")[$sin ( θ ) = display(frac(15, 17)) , " " " " tan ( θ ) = display(frac(− 15, 8))$] === Using a Calculator In the Examples above, we used a point on the terminal side to find exact values for the trigonometric ratios of obtuse angles. Scientific and graphing calculators are programmed with approximations for these trig ratios. #examplebox("Example 3")[][ Find the sine and cosine of #math.equation(block: false, alt: "130 degrees")[$130^(∘)$]. Compare to the sine and cosine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]. #solutionbox[ Using a calculator and rounding the values to four places, we find #math.equation(block: true, alt: "sin ⁡ open parenthesis 130 degrees close parenthesis, equals 0.7660 and cos ⁡ open parenthesis 130 degrees close parenthesis equals minus 0.6428; sin ⁡ open parenthesis 50 degrees close parenthesis, equals 0.7660 and cos ⁡ open parenthesis 50 degrees close parenthesis equals 0.6428")[$sin ( 130^(∘) ) & = 0.7660 " " " " " " " " " " "and" " " " " " " " " " " cos ( 130^(∘) ) = − 0.6428 \ sin ( 50^(∘) ) & = 0.7660 " " " " " " " " " " "and" " " " " " " " " " " " " " " cos ( 50^(∘) ) = 0.6428$] We see that #math.equation(block: false, alt: "sin ⁡ open parenthesis 130 degrees close parenthesis equals sin ⁡ open parenthesis 50 degrees close parenthesis")[$sin ( 130^(∘) ) = sin ( 50^(∘) )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 130 degrees close parenthesis equals minus cos ⁡ open parenthesis 50 degrees close parenthesis")[$cos ( 130^(∘) ) = − cos ( 50^(∘) )$]. This result should not be surprising when we look at both angles in standard position, as shown below. The angles #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] and #math.equation(block: false, alt: "130 degrees")[$130^(∘)$] are supplementary. The right triangles formed by choosing the points #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] and #math.equation(block: false, alt: "open parenthesis minus x , y close parenthesis")[$( − x , y )$] on their terminal sides are congruent triangles. #figure(figph[supplementary angles], alt: "supplementary angles", caption: none) Consequently, the trigonometric ratios for #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] and for #math.equation(block: false, alt: "130 degrees")[$130^(∘)$] are equal, except that the cosine of #math.equation(block: false, alt: "130 degrees")[$130^(∘)$] is negative. ] ] Use your calculator to fill in the table. Round to four decimal places. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " " " cos ( θ ) " " " " " " " "$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$" " " " " " " " sin ( θ ) " " " " " " " "$]], [#math.equation(block: false, alt: "180 degrees minus θ")[$180^(∘) − θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$cos ( 180^(∘) − θ )$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$sin ( 180^(∘) − θ )$]]), [#math.equation(block: false, alt: "10 degrees")[$10^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "20 degrees")[$20^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "40 degrees")[$40^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "50 degrees")[$50^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "70 degrees")[$70^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "80 degrees")[$80^(∘)$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " " " cos ( θ ) " " " " " " " "$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$" " " " " " " " sin ( θ ) " " " " " " " "$]], [#math.equation(block: false, alt: "180 degrees minus θ")[$180^(∘) − θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$cos ( 180^(∘) − θ )$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$sin ( 180^(∘) − θ )$]]), [#math.equation(block: false, alt: "10 degrees")[$10^(∘)$]], [#math.equation(block: false, alt: "0.9848")[$0.9848$]], [#math.equation(block: false, alt: "0.1736")[$0.1736$]], [#math.equation(block: false, alt: "170 degrees")[$170^(∘)$]], [#math.equation(block: false, alt: "minus 0.9848")[$− 0.9848$]], [#math.equation(block: false, alt: "0.1736")[$0.1736$]], [#math.equation(block: false, alt: "20 degrees")[$20^(∘)$]], [#math.equation(block: false, alt: "0.9397")[$0.9397$]], [#math.equation(block: false, alt: "0.3420")[$0.3420$]], [#math.equation(block: false, alt: "160 degrees")[$160^(∘)$]], [#math.equation(block: false, alt: "minus 0.9397")[$− 0.9397$]], [#math.equation(block: false, alt: "0.3420")[$0.3420$]], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "0.8660")[$0.8660$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "150 degrees")[$150^(∘)$]], [#math.equation(block: false, alt: "0.8660")[$0.8660$]], [#math.equation(block: false, alt: "minus 0.5")[$− 0.5$]], [#math.equation(block: false, alt: "40 degrees")[$40^(∘)$]], [#math.equation(block: false, alt: "0.7660")[$0.7660$]], [#math.equation(block: false, alt: "0.6428")[$0.6428$]], [#math.equation(block: false, alt: "140 degrees")[$140^(∘)$]], [#math.equation(block: false, alt: "minus 0.7660")[$− 0.7660$]], [#math.equation(block: false, alt: "0.6428")[$0.6428$]], [#math.equation(block: false, alt: "50 degrees")[$50^(∘)$]], [#math.equation(block: false, alt: "0.6428")[$0.6428$]], [#math.equation(block: false, alt: "0.7660")[$0.7660$]], [#math.equation(block: false, alt: "130 degrees")[$130^(∘)$]], [#math.equation(block: false, alt: "minus 0.6428")[$− 0.6428$]], [#math.equation(block: false, alt: "0.7660")[$0.7660$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.8660")[$0.8660$]], [#math.equation(block: false, alt: "120 degrees")[$120^(∘)$]], [#math.equation(block: false, alt: "minus 0.5")[$− 0.5$]], [#math.equation(block: false, alt: "0.8660")[$0.8660$]], [#math.equation(block: false, alt: "70 degrees")[$70^(∘)$]], [#math.equation(block: false, alt: "0.3420")[$0.3420$]], [#math.equation(block: false, alt: "0.9397")[$0.9397$]], [#math.equation(block: false, alt: "110 degrees")[$110^(∘)$]], [#math.equation(block: false, alt: "minus 0.9397")[$− 0.9397$]], [#math.equation(block: false, alt: "0.3420")[$0.3420$]], [#math.equation(block: false, alt: "80 degrees")[$80^(∘)$]], [#math.equation(block: false, alt: "0.1736")[$0.1736$]], [#math.equation(block: false, alt: "0.9848")[$0.9848$]], [#math.equation(block: false, alt: "100 degrees")[$100^(∘)$]], [#math.equation(block: false, alt: "minus 0.9848")[$− 0.9848$]], [#math.equation(block: false, alt: "0.1736")[$0.1736$]], )) === Trigonometric Ratios for Supplementary Angles The Examples above illustrate the following equations for supplementary angles. These three equations are called #strong[identities], which means that they are true for all values of the variable #math.equation(block: false, alt: "θ")[$θ$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Trigonometric Ratios for Supplementary Angles] - #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis equals minus cos ⁡ open parenthesis θ close parenthesis")[$cos ( 180^(∘) − θ ) = − cos ( θ )$] - #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$sin ( 180^(∘) − θ ) = sin ( θ )$] - #math.equation(block: false, alt: "tan ⁡ open parenthesis 180 degrees minus θ close parenthesis equals minus tan ⁡ open parenthesis θ close parenthesis")[$tan ( 180^(∘) − θ ) = − tan ( θ )$] #figure(figph[supplementary angles in standard position], alt: "supplementary angles in standard position", caption: none) ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Because of these relationships, there are always two (supplementary) angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] that have the same sine. Your calculator will only tell you one of them, so you have to be able to find the other one on your own! Fortunately, this is not difficult. ] #examplebox("Example 4")[][ Find two different angles #math.equation(block: false, alt: "θ")[$θ$], rounded to the nearest #math.equation(block: false, alt: "0.1 degrees")[$0.1^(∘)$], that satisfy #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.25")[$" " sin ( θ ) = 0.25$]. #solutionbox[ To find an angle with #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.25")[$" " sin ( θ ) = 0.25$], we calculate #math.equation(block: false, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis 0.25 close parenthesis")[$" " θ = sin^(− 1) ( 0.25 )$]. With the calculator in degree mode, we press $#h(2em) #h(2em) #h(2em)$2nd SIN 0.25 ENTER and find that one angle is #math.equation(block: false, alt: "θ approximately equals 14.5 degrees")[$" " θ ≈ 14.5^(∘)$]. We draw this acute angle in standard position in the first quadrant, and sketch in a right triangle as shown below. There must also be an obtuse angle whose sine is #math.equation(block: false, alt: "0.25")[$0.25$]. To see the second angle, we draw a congruent triangle in the second quadrant as shown. #figure(figph[supplementary angle], alt: "supplementary angle", caption: none) The supplement of #math.equation(block: false, alt: "14.5 degrees")[$14.5^(∘)$], namely #math.equation(block: false, alt: "θ equals 180 degrees minus 14.5 degrees equals 165.5 degrees")[$" " θ = 180^(∘) − 14.5^(∘) = 165.5^(∘)$], is the obtuse angle we need. Because the ratio #math.equation(block: false, alt: "the fraction y over r")[$display(frac(y, r))$] is the same for both triangles, they have the same sine. ] ] Find two different angles #math.equation(block: false, alt: "θ")[$θ$] that satisfy #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.8")[$" " sin ( θ ) = 0.8$]. One angle is #math.equation(block: false, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis 0.8 close parenthesis equals 53.13 degrees")[$" " θ = sin^(− 1) ( 0.8 ) = 53.13^(∘)$]. The second angle is the supplement of #math.equation(block: false, alt: "53.13 degrees")[$53.13^(∘)$], or #math.equation(block: false, alt: "open parenthesis 180 minus 53.13 close parenthesis equals 126.87 degrees")[$" " ( 180 − 53.13 ) = 126.87^(∘)$]. Because there are two angles with the same sine, it is easier to find an obtuse angle if we know its cosine instead of its sine. #examplebox("Example 5")[][ Find the angle shown at right. #figure(figph[obtuse angle], alt: "obtuse angle", caption: none) #solutionbox[ Using #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$] and #math.equation(block: false, alt: "y equals 4")[$y = 4$], we find #math.equation(block: true, alt: "r equals the square root of 3 squared plus 4 squared equals the square root of 25 equals 5")[$r = sqrt(3^(2) + 4^(2)) = sqrt(25) = 5$] so #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r equals the fraction minus 3 over 5")[$cos ( θ ) = display(frac(x, r)) = display(frac(− 3, 5))$], and #math.equation(block: false, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis the fraction minus 3 over 5 close parenthesis .")[$θ = cos^(− 1) ( display(frac(− 3, 5)) ) .$] We can enter $#h(2em) #h(2em) #h(2em)$ 2nd COS -3 5 ENTER to see that #math.equation(block: false, alt: "θ approximately equals 126.9 degrees")[$θ ≈ 126.9^(∘)$]. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the previous example, you might notice that #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 4 over 3")[$" " tan ( θ ) = display(frac(− 4, 3))$] and try to find by calculating #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis the fraction minus 4 over 3 close parenthesis")[$" " tan^(− 1) ( display(frac(− 4, 3)) )$]. However, if we press $#h(2em) #h(2em) #h(2em)$2nd TAN 4 3 ENTER the calculator returns an angle of #math.equation(block: false, alt: "θ approximately equals minus 53.1 degrees")[$" " θ ≈ − 53.1^(∘)$]. It is true that #math.equation(block: false, alt: "tan ⁡ open parenthesis minus 53.1 degrees close parenthesis equals the fraction minus 4 over 3")[$" " tan ( − 53.1^(∘) ) = display(frac(− 4, 3))$], but this is not the obtuse angle we want. We also know that #math.equation(block: false, alt: "sin ⁡ θ equals the fraction 4 over 5")[$" " sin θ = display(frac(4, 5))$], and if we press $#h(2em)$2nd SIN 4 5 ENTER we get #math.equation(block: false, alt: "θ approximately equals 53.1 degrees")[$θ ≈ 53.1^(∘)$]. #figure(figph[supplementary angles], alt: "supplementary angles", caption: none) This is the acute angle whose terminal side passes through the point #math.equation(block: false, alt: "open parenthesis 3 , 4 close parenthesis")[$( 3 , 4 )$], as shown in the figure above. The angle we want is its supplement, #math.equation(block: false, alt: "θ approximately equals 180 degrees minus 53.1 degrees equals 126.9 degrees")[$" " θ ≈ 180^(∘) − 53.1^(∘) = 126.9^(∘)$]. ] + Find the cosine of an obtuse angle with #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 2")[$tan ( θ ) = − 2$] . + Find the angle #math.equation(block: false, alt: "θ")[$θ$] in part (a). + The point #math.equation(block: false, alt: "open parenthesis minus 1 , 2 close parenthesis")[$( − 1 , 2 )$] lies on the terminal side of the angle, so #math.equation(block: true, alt: "r squared equals open parenthesis minus 1 close parenthesis squared plus 2 squared equals 5")[$r^(2) = ( − 1 )^(2) + 2^(2) = 5$] and #math.equation(block: false, alt: "r equals the square root of 5")[$r = sqrt(5)$]. Then #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over the square root of 5")[$" " cos ( θ ) = display(frac(− 1, sqrt(5)))$]. + #math.equation(block: false, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis the fraction minus 1 over the square root of 5 close parenthesis approximately equals 116.565 degrees")[$θ = cos^(− 1) ( display(frac(− 1, sqrt(5))) ) ≈ 116.565^(∘)$] === Supplements of the Special Angles In Chapter 2 we learned that the angles #math.equation(block: false, alt: "30 degrees , 45 degrees")[$30^(∘) , 45^(∘)$] and #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] are useful because we can find exact values for their trigonometric ratios. The same is true for the supplements of these angles in the second quadrant, shown at right. #figure(figph[supplements of special angles], alt: "supplements of special angles", caption: none) #examplebox("Example 6")[][ Find exact values for the trigonometric ratios of #math.equation(block: false, alt: "135 degrees")[$135^(∘)$]. #solutionbox[ We sketch an angle of #math.equation(block: false, alt: "θ equals 135 degrees")[$θ = 135^(∘)$] in standard position, as shown below. The terminal side is in the second quadrant and makes an acute angle of #math.equation(block: false, alt: "45 degrees")[$45^(∘)$] with the negative #math.equation(block: false, alt: "x")[$x$]-axis, and passes through the point #math.equation(block: false, alt: "open parenthesis minus 1 , 1 close parenthesis")[$( − 1 , 1 )$]. Thus, #math.equation(block: false, alt: "r equals the square root of open parenthesis minus 1 close parenthesis squared plus 1 squared equals the square root of 2")[$" " r = sqrt(( − 1 )^(2) + 1^(2)) = sqrt(2) " "$], and we calculate #math.equation(block: true, alt: "cos ⁡ open parenthesis 135 degrees close parenthesis, equals the fraction x over r equals the fraction minus 1 over the square root of 2; sin ⁡ open parenthesis 135 degrees close parenthesis, equals the fraction y over r equals the fraction 1 over the square root of 2; tan ⁡ open parenthesis 135 degrees close parenthesis, equals the fraction y over x equals the fraction 1 over minus 1 equals minus 1")[$cos ( 135^(∘) ) & = display(frac(x, r)) = display(frac(− 1, sqrt(2))) \ sin ( 135^(∘) ) & = display(frac(y, r)) = display(frac(1, sqrt(2))) \ tan ( 135^(∘) ) & = display(frac(y, x)) = display(frac(1, − 1)) = − 1$] #figure(figph[obtuse angle 135 degrees], alt: "obtuse angle 135 degrees", caption: none) ] ] Find exact values for the trigonometric ratios of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] and #math.equation(block: false, alt: "150 degrees")[$150^(∘)$]. #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] is the supplement of #math.equation(block: false, alt: "60 degrees")[$60^(∘)$], and #math.equation(block: false, alt: "150 degrees")[$150^(∘)$] is the supplement of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "120 degrees")[$120^(∘)$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "150 degrees")[$150^(∘)$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 3")[$display(frac(− 1, sqrt(3)))$]], )) We can also find the trig ratios for the #strong[quadrantal] angles. These are the angles, including #math.equation(block: false, alt: "0 degrees")[$0^(∘)$], #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], whose terminal sides lie on one of the axes. #examplebox("Example 7")[][ Find exact values for the trigonometric ratios of #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. #solutionbox[ The terminal side of a #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle in standard position is the positive #math.equation(block: false, alt: "y")[$y$]-axis. If we take the point #math.equation(block: false, alt: "P open parenthesis 0 , 1 close parenthesis")[$P ( 0 , 1 )$] on the terminal side as shown at right, then #math.equation(block: false, alt: "x equals 0")[$x = 0$] and #math.equation(block: false, alt: "y equals 1")[$y = 1$]. Although we don't have a triangle, we can still calculate a value for #math.equation(block: false, alt: "r")[$r$], the distance from the origin to #math.equation(block: false, alt: "P")[$P$]. #math.equation(block: true, alt: "r equals the square root of 0 squared plus 1 squared equals 1")[$r = sqrt(0^(2) + 1^(2)) = 1$] #figure(figph[right angle], alt: "right angle", caption: none) Our coordinate definitions for the trig ratios give us #math.equation(block: true, alt: "cos ⁡ open parenthesis 90 degrees close parenthesis equals the fraction x over r equals the fraction 0 over 1 and sin ⁡ open parenthesis 90 degrees close parenthesis equals the fraction y over r equals the fraction 1 over 1")[$cos ( 90^(∘) ) = display(frac(x, r)) = display(frac(0, 1)) " " " " " " " " " " "and" " " " " " " " " " " sin ( 90^(∘) ) = display(frac(y, r)) = display(frac(1, 1))$] so #math.equation(block: false, alt: "cos ⁡ open parenthesis 90 degrees close parenthesis equals 0")[$" " cos ( 90^(∘) ) = 0 " "$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 90 degrees close parenthesis equals 1.")[$" " sin ( 90^(∘) ) = 1 " " .$] Also, #math.equation(block: false, alt: "tan ⁡ open parenthesis 90 degrees close parenthesis equals the fraction y over x equals the fraction 1 over 0 ,")[$" " tan ( 90^(∘) ) = display(frac(y, x)) = display(frac(1, 0)) " " ,$] so #math.equation(block: false, alt: "tan ⁡ open parenthesis 90 degrees close parenthesis")[$tan ( 90^(∘) )$] is undefined. ] ] Find exact values for the trigonometric ratios of #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. The point #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] lies on the terminal side of the angle #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. Thus #math.equation(block: true, alt: "cos ⁡ open parenthesis 180 degrees close parenthesis equals minus 1 , sin ⁡ open parenthesis 180 degrees close parenthesis equals 0 , tan ⁡ open parenthesis 180 degrees close parenthesis equals 0")[$cos ( 180^(∘) ) = − 1 , " " sin ( 180^(∘) ) = 0 , " " tan ( 180^(∘) ) = 0$] === The Area of a Triangle The figure below shows part of the map for a new housing development, Pacific Shores. You are interested in the corner lot, number 86, and you would like to know the area of the lot in square feet. The sales representative for Pacific Shores provides you with the dimensions of the lot, but you don't know a formula for the area of an irregularly shaped quadrilateral. #figure(figph[map of house lots], alt: "map of house lots", caption: none) It occurs to you that you can divide the quadrilateral into two triangles, and find the area of each. Now, you know a formula for the area of a triangle in terms of its base and height, namely, #math.equation(block: true, alt: "A equals the fraction 1 over 2 b h")[$A = display(frac(1, 2)) b h$] , #figure(figph[Lot 86 quadrilateral], alt: "Lot 86 quadrilateral", caption: none) but unfortunately, you don't know the height of either triangle. However, you can easily measure the angles at the corners of the lot using the plot map and a protractor. You can check the values on the plot map for lot 86 shown above. Using trigonometry, we can find the area of a triangle if we know two of its sides, say #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], and the included angle, #math.equation(block: false, alt: "θ")[$θ$]. The figure below shows three possibilities, depending on whether the angle #math.equation(block: false, alt: "θ")[$θ$] is acute, obtuse, or #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. #figure(figph[three triangles with altitudes], alt: "three triangles with altitudes", caption: none) In each case, #math.equation(block: false, alt: "b")[$b$] is the base of the triangle, and its altitude is #math.equation(block: false, alt: "h")[$h$]. Our task is to find an expression for #math.equation(block: false, alt: "h")[$h$] in terms of the quantities we know: #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "θ")[$θ$]. You should check that in all three triangles #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction h over a")[$sin ( θ ) = display(frac(h, a))$] Solving for #math.equation(block: false, alt: "h")[$h$] gives us #math.equation(block: false, alt: "h equals a sin ⁡ open parenthesis θ close parenthesis")[$h = a sin ( θ )$]. Finally, we substitute this expression for #math.equation(block: false, alt: "h")[$h$] into our old formula for the area to get #math.equation(block: true, alt: "A equals the fraction 1 over 2 b h equals the fraction 1 over 2 b a sin ⁡ open parenthesis θ close parenthesis")[$A = display(frac(1, 2)) b " " bold(italic(h)) = display(frac(1, 2)) b " " bold(italic(a)) bold(sin) ( bold(italic(θ)) )$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Area of a Triangle] If a triangle has sides of length #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], and the angle between those two sides is #math.equation(block: false, alt: "θ")[$θ$], then the area of the triangle is given by #math.equation(block: true, alt: "A equals the fraction 1 over 2 a b sin ⁡ open parenthesis θ close parenthesis")[$bold(italic(A)) = display(frac(1, 2)) bold(italic(a)) bold(italic(b)) bold(sin) ( bold(italic(θ)) )$] #figure(figph[Triangle-angle-theta-sides-a-and-b], alt: "Triangle-angle-theta-sides-a-and-b", caption: none) ] #examplebox("Example 8")[][ Find the area of lot 86. #solutionbox[ For the triangle in the lower portion of lot 86, #math.equation(block: false, alt: "a equals 120.3")[$a = 120.3$], #math.equation(block: false, alt: "b equals 141")[$b = 141$], and #math.equation(block: false, alt: "θ equals 95 degrees")[$θ = 95^(∘)$]. The area of that portion is #math.equation(block: true, alt: "First Area, equals the fraction 1 over 2 a b sin ⁡ open parenthesis θ close parenthesis; equals the fraction 1 over 2 open parenthesis 120.3 close parenthesis open parenthesis open parenthesis 141 close parenthesis sin ⁡ open parenthesis 95 degrees close parenthesis approximately equals 8448.88")[$"First Area" & = display(frac(1, 2)) a b sin \( θ \) \ & = display(frac(1, 2)) \( 120.3 \) \( \( 141 \) " " sin \( 95^(∘) \) ≈ 8448.88$] For the triangle in the upper portion of the lot, #math.equation(block: false, alt: "a equals 161")[$a = 161$], #math.equation(block: false, alt: "b equals 114.8")[$b = 114.8$], and #math.equation(block: false, alt: "θ equals 86.1 degrees")[$θ = 86.1^(∘)$]. The area of that portion is #math.equation(block: true, alt: "Second Area, equals the fraction 1 over 2 a b sin ⁡ open parenthesis θ close parenthesis; equals the fraction 1 over 2 open parenthesis 161 close parenthesis open parenthesis open parenthesis 114.8 close parenthesis sin ⁡ open parenthesis 86.1 degrees close parenthesis approximately equals 9220.00")[$"Second Area" & = display(frac(1, 2)) a b sin \( θ \) \ & = display(frac(1, 2)) \( 161 \) \( \( 114.8 \) " " sin \( 86.1^(∘) \) ≈ 9220.00$] The total area of the lot is the sum of the areas of the triangles #math.equation(block: true, alt: "Total area equals First Area plus Second Area approximately equals 17668.88")[$"Total area" = "First Area" + "Second Area" ≈ 17668.88$] Lot 86 has an area of approximately 17,669 square feet. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 a b sin ⁡ open parenthesis θ close parenthesis")[$" " A = display(frac(1, 2)) a b sin ( θ ) " "$] does #emph[not] mean that we always use the sides labeled #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] to find the area of a triangle. In this formula, the variables #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] represent the lengths of the sides that #emph[include] the known angle. For example, the area of the triangle at right is given by #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis 5 c close parenthesis sin ⁡ open parenthesis ϕ close parenthesis")[$A = display(frac(1, 2)) ( 5 c ) sin ( ϕ )$]. #figure(figph[triangle], alt: "triangle", caption: none) ] A triangle has sides of length 6 and 7, and the angle between those sides is #math.equation(block: false, alt: "150 degrees")[$150^(∘)$]. Find the area of the triangle. The area is given by #math.equation(block: true, alt: "A equals the fraction 1 over 2 open parenthesis 6 close parenthesis open parenthesis 7 close parenthesis sin ⁡ open parenthesis 150 degrees close parenthesis equals 21 open parenthesis the fraction 1 over 2 close parenthesis equals the fraction 21 over 2")[$A = display(frac(1, 2)) ( 6 ) ( 7 ) sin ( 150^(∘) ) = 21 ( display(frac(1, 2)) ) = display(frac(21, 2))$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the area of the triangle. 1. #figure(figph[triangle], alt: "triangle", caption: none) 2. #figure(figph[triangle], alt: "triangle", caption: none) 3. #figure(figph[triangle], alt: "triangle", caption: none) 4. #figure(figph[triangle], alt: "triangle", caption: none) How many degrees are in each fraction of one complete revolution? 5. #math.equation(block: false, alt: "the fraction 1 over 4")[$#h(0.222em) display(frac(1, 4))$] 6. #math.equation(block: false, alt: "the fraction 1 over 5")[$#h(0.222em) display(frac(1, 5))$] 7. #math.equation(block: false, alt: "the fraction 1 over 6")[$#h(0.222em) display(frac(1, 6))$] 8. #math.equation(block: false, alt: "the fraction 1 over 8")[$#h(0.222em) display(frac(1, 8))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "24")[$24$] + #math.equation(block: false, alt: "24")[$24$] + #math.equation(block: false, alt: "24")[$24$] + #math.equation(block: false, alt: "24")[$24$] + #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] + #math.equation(block: false, alt: "72 degrees")[$72^(∘)$] + #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] + #math.equation(block: false, alt: "45 degrees")[$45^(∘)$] ] === Section 3.1 Summary ==== Vocabulary - Standard position - Initial side - Terminal side - Quadrantal angle - Oblique triangle - Quadrilateral - Identity ==== Concepts + We put an angle in #strong[standard position] by placing its vertex at the origin and the #strong[initial side] on the positive #math.equation(block: false, alt: "x")[$x$]-axis. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Coordinate Definitions of the Trigonometric Ratios] - #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r")[$cos ( θ ) = display(frac(x, r))$] - #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r")[$sin ( θ ) = display(frac(y, r))$] - #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x")[$tan ( θ ) = display(frac(y, x))$] #figure(figph[supplementary angles in standard position], alt: "supplementary angles in standard position", caption: none) ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Trigonometric Ratios for Supplementary Angles] - #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis equals minus cos ⁡ θ")[$cos ( 180^(∘) − θ ) = − cos θ$] - #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis equals sin ⁡ θ")[$sin ( 180^(∘) − θ ) = sin θ$] - #math.equation(block: false, alt: "tan ⁡ open parenthesis 180 degrees minus θ close parenthesis equals minus tan ⁡ θ")[$tan ( 180^(∘) − θ ) = − tan θ$] #figure(figph[supplementary angles in standard position], alt: "supplementary angles in standard position", caption: none) ] + There are always two (supplementary) angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] that have the same sine. Your calculator will only tell you one of them. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Area of a Triangle] If a triangle has sides of length #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], and the angle between those two sides is #math.equation(block: false, alt: "θ")[$θ$], then the area of the triangle is given by #math.equation(block: true, alt: "A equals the fraction 1 over 2 a b sin ⁡ open parenthesis θ close parenthesis")[$A = display(frac(1, 2)) a b sin ( θ )$]#figure(figph[Triangle-angle-theta-sides-a-and-b], alt: "Triangle-angle-theta-sides-a-and-b", caption: none) ] ==== Study Questions + Delbert says that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 7")[$sin ( θ ) = display(frac(4, 7))$] in the figure. Is he correct? Why or why not?#figure(figph[triangle], alt: "triangle", caption: none) + Give the lengths of the legs of each right triangle.a. #figure(figph[triangle], alt: "triangle", caption: none) b. #figure(figph[triangle], alt: "triangle", caption: none) + Explain why the length of the horizontal leg of the right triangle is #math.equation(block: false, alt: "minus x")[$− x$] .#figure(figph[triangle], alt: "triangle", caption: none) + Why are the sines of supplementary angles equal, but the cosines are not? What about the tangents of supplementary angles? + Use your calculator to evaluate #math.equation(block: false, alt: "sin ⁡ open parenthesis 118 degrees close parenthesis")[$sin ( 118^(∘) )$], then evaluate #math.equation(block: false, alt: "sin to the power minus 1 ⁡ (ANS)")[$sin^(− 1) "(ANS)"$] . Explain the result. + Write an expression for the area of the triangle. #figure(figph[triangle], alt: "triangle", caption: none) ==== Skills Practice each skill in the Homework Problems listed. + Use the coordinate definition of the trig ratios \#3-20, 45-48 + Find the trig ratios of supplementary angles \#7-10, 21-38 + Know the trig ratios of the special angles in the second quadrant \#21, 41-44 + Find two solutions of the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals k")[$sin ( θ ) = k$] \#29-38 + Find the area of a triangle \#49-58 === Homework 3.1 Without using pencil and paper or a calculator, give the supplement of each angle. + #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] + #math.equation(block: false, alt: "45 degrees")[$45^(∘)$] + #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] + #math.equation(block: false, alt: "25 degrees")[$25^(∘)$] + #math.equation(block: false, alt: "165 degrees")[$165^(∘)$] + #math.equation(block: false, alt: "110 degrees")[$110^(∘)$] + #math.equation(block: false, alt: "150 degrees")[$150^(∘)$] + #math.equation(block: false, alt: "135 degrees")[$135^(∘)$] + #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] + #math.equation(block: false, alt: "155 degrees")[$155^(∘)$] + #math.equation(block: false, alt: "15 degrees")[$15^(∘)$] + #math.equation(block: false, alt: "70 degrees")[$70^(∘)$] Without using pencil and paper or a calculator, give the complement of each angle. + #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] + #math.equation(block: false, alt: "80 degrees")[$80^(∘)$] + #math.equation(block: false, alt: "25 degrees")[$25^(∘)$] + #math.equation(block: false, alt: "18 degrees")[$18^(∘)$] + #math.equation(block: false, alt: "64 degrees")[$64^(∘)$] + #math.equation(block: false, alt: "47 degrees")[$47^(∘)$] For Problems 3–6, + Give the coordinates of point #math.equation(block: false, alt: "P")[$P$] on the terminal side of the angle. + Find the distance from the origin to point #math.equation(block: false, alt: "P")[$P$]. + Find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis ,")[$cos ( θ ) , " " sin ( θ ) ,$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis .")[$" " tan ( θ ) .$] #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "open parenthesis 5 , 2 close parenthesis")[$( 5 , 2 )$] + #math.equation(block: false, alt: "the square root of 29")[$sqrt(29)$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 5 over the square root of 29 , sin ⁡ open parenthesis θ close parenthesis equals the fraction 2 over the square root of 29 , tan ⁡ open parenthesis θ close parenthesis equals the fraction 2 over 5")[$cos ( θ ) = display(frac(5, sqrt(29))) , " " " " sin ( θ ) = display(frac(2, sqrt(29))) , " " " " tan ( θ ) = display(frac(2, 5))$] #figure(figph[angle], alt: "angle", caption: none) #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 4 , 7 close parenthesis")[$( − 4 , 7 )$] + #math.equation(block: false, alt: "the square root of 65")[$sqrt(65)$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 4 over the square root of 65 , sin ⁡ open parenthesis θ close parenthesis equals the fraction 7 over the square root of 65 , tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 7 over 4")[$cos ( θ ) = display(frac(− 4, sqrt(65))) , " " " " sin ( θ ) = display(frac(7, sqrt(65))) , " " " " tan ( θ ) = display(frac(− 7, 4))$] #figure(figph[angle], alt: "angle", caption: none) For Problems 7–10, + Find the sine and cosine of the angle. + Sketch the supplement of the angle in standard position. (Use congruent triangles.) + Find the sine and cosine of the supplement. + Find the angle and its supplement, rounded to the nearest degree. #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 9 over the square root of 97")[$sin ( θ ) = display(frac(9, sqrt(97)))$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 4 over the square root of 97")[$" " cos ( θ ) = display(frac(4, sqrt(97)))$] + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis equals the fraction 9 over the square root of 97")[$sin ( 180^(∘) − θ ) = display(frac(9, sqrt(97)))$], #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis equals the fraction minus 4 over the square root of 97")[$" " cos ( 180^(∘) − θ ) = display(frac(− 4, sqrt(97)))$] + #math.equation(block: false, alt: "θ equals 66 degrees , 180 degrees minus θ equals 114 degrees")[$θ = 66^(∘) , " " " " 180^(∘) − θ = 114^(∘)$] #figure(figph[angle], alt: "angle", caption: none) #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 8 over the square root of 89")[$sin ( θ ) = display(frac(8, sqrt(89)))$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over the square root of 89")[$" " cos ( θ ) = display(frac(− 5, sqrt(89)))$] + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis equals the fraction 8 over the square root of 89")[$sin ( 180^(∘) − θ ) = display(frac(8, sqrt(89)))$], #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis equals the fraction 5 over the square root of 89")[$" " cos ( 180^(∘) − θ ) = display(frac(5, sqrt(89)))$] + #math.equation(block: false, alt: "θ equals 122 degrees , 180 degrees minus θ equals 58 degrees")[$θ = 122^(∘) , " " " " 180^(∘) − θ = 58^(∘)$] #figure(figph[angle], alt: "angle", caption: none) For Problems 11–20, + Sketch an angle in standard position with the given properties. + Find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis ,")[$cos ( θ ) , " " " " sin ( θ ) ,$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis .")[$tan ( θ ) .$] + Find the angle #math.equation(block: false, alt: "θ")[$θ$], rounded to tenths of a degree. The point #math.equation(block: false, alt: "open parenthesis minus 5 , 12 close parenthesis")[$( − 5 , 12 )$] is on the terminal side. #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over 13")[$cos ( θ ) = display(frac(− 5, 13))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 12 over 13")[$" " sin ( θ ) = display(frac(12, 13))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 12 over 5")[$" " tan ( θ ) = display(frac(− 12, 5))$] + #math.equation(block: false, alt: "112.6 degrees")[$112.6^(∘)$] The point #math.equation(block: false, alt: "open parenthesis 12 , 9 close parenthesis")[$( 12 , 9 )$] is on the terminal side. #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) #math.equation(block: true, alt: "cos ⁡ θ equals minus 0.8")[$cos θ = − 0.8$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5 , tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 4")[$cos ( θ ) = display(frac(3, 5)) , " " " " tan ( θ ) = display(frac(− 3, 4))$] + #math.equation(block: false, alt: "143.1 degrees")[$143.1^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 13")[$cos ( θ ) = display(frac(5, 13))$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 11")[$cos ( θ ) = display(frac(3, 11))$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 112 over 11")[$sin ( θ ) = display(frac(sqrt(112), 11))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 112 over 3")[$" " tan ( θ ) = display(frac(sqrt(112), 3))$] + #math.equation(block: false, alt: "74.2 degrees")[$74.2^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over 6")[$cos ( θ ) = display(frac(− 5, 6))$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over 6")[$tan ( θ ) = display(frac(− 1, 6))$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over the square root of 37")[$sin ( θ ) = display(frac(1, sqrt(37)))$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 6 over the square root of 37")[$" " cos ( θ ) = display(frac(− 6, sqrt(37)))$] + #math.equation(block: false, alt: "170.5 degrees")[$170.5^(∘)$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 9 over 5")[$tan ( θ ) = display(frac(9, 5))$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals 4")[$tan ( θ ) = 4$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) + #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 4 over the square root of 17")[$sin ( θ ) = display(frac(4, sqrt(17)))$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 1 over the square root of 17")[$" " cos ( θ ) = display(frac(1, sqrt(17)))$] + #math.equation(block: false, alt: "76.0 degrees")[$76.0^(∘)$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 1")[$tan ( θ ) = − 1$] #figure(figph[Grid for quadrants I and II], alt: "Grid for quadrants I and II", caption: none) Fill in exact values from memory without using a calculator. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0 degrees")[$" " " " " " 0^(∘) " " " " " "$]], [#math.equation(block: false, alt: "30 degrees")[$" " " " " " 30^(∘) " " " " " "$]], [#math.equation(block: false, alt: "45 degrees")[$" " " " " " 45^(∘) " " " " " "$]], [#math.equation(block: false, alt: "60 degrees")[$" " " " " " 60^(∘) " " " " " "$]], [#math.equation(block: false, alt: "90 degrees")[$" " " " " " 90^(∘) " " " " " "$]], [#math.equation(block: false, alt: "120 degrees")[$" " " " " " 120^(∘) " " " " " "$]], [#math.equation(block: false, alt: "135 degrees")[$" " " " " " 135^(∘) " " " " " "$]], [#math.equation(block: false, alt: "150 degrees")[$" " " " " " 150^(∘) " " " " " "$]], [#math.equation(block: false, alt: "180 degrees")[$" " " " " " 180^(∘) " " " " " "$]]), [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], )) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0 degrees")[$" " " " " " 0^(∘) " " " " " "$]], [#math.equation(block: false, alt: "30 degrees")[$" " " " " " 30^(∘) " " " " " "$]], [#math.equation(block: false, alt: "45 degrees")[$" " " " " " 45^(∘) " " " " " "$]], [#math.equation(block: false, alt: "60 degrees")[$" " " " " " 60^(∘) " " " " " "$]], [#math.equation(block: false, alt: "90 degrees")[$" " " " " " 90^(∘) " " " " " "$]], [#math.equation(block: false, alt: "120 degrees")[$" " " " " " 120^(∘) " " " " " "$]], [#math.equation(block: false, alt: "135 degrees")[$" " " " " " 135^(∘) " " " " " "$]], [#math.equation(block: false, alt: "150 degrees")[$" " " " " " 150^(∘) " " " " " "$]], [#math.equation(block: false, alt: "180 degrees")[$" " " " " " 180^(∘) " " " " " "$]]), [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2))) " "$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "undefined")[$"undefined"$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 3")[$display(frac(− 1, sqrt(3)))$]], [#math.equation(block: false, alt: "0")[$0$]], )) Use your calculator to fill in the table. Round values to four decimal places. #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "15 degrees")[$" " " " " " 15^(∘) " " " " " "$]], [#math.equation(block: false, alt: "25 degrees")[$" " " " " " 25^(∘) " " " " " "$]], [#math.equation(block: false, alt: "65 degrees")[$" " " " " " 65^(∘) " " " " " "$]], [#math.equation(block: false, alt: "75 degrees")[$" " " " " " 75^(∘) " " " " " "$]], [#math.equation(block: false, alt: "105 degrees")[$" " " " " " 105^(∘) " " " " " "$]], [#math.equation(block: false, alt: "115 degrees")[$" " " " " " 115^(∘) " " " " " "$]], [#math.equation(block: false, alt: "155 degrees")[$" " " " " " 155^(∘) " " " " " "$]], [#math.equation(block: false, alt: "165 degrees")[$" " " " " " 165^(∘) " " " " " "$]]), [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], )) For each angle #math.equation(block: false, alt: "θ")[$θ$] in the table for Problem 22, the angle #math.equation(block: false, alt: "180 degrees minus θ")[$180^(∘) − θ$] is also in the table. + What is true about #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$sin ( 180^(∘) − θ )$]? + What is true about #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$cos ( 180^(∘) − θ )$]? + What is true about #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$tan ( 180^(∘) − θ )$]? + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$sin ( θ ) = sin ( 180^(∘) − θ )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus cos ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$cos ( θ ) = − cos ( 180^(∘) − θ )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus tan ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$tan ( θ ) = − tan ( 180^(∘) − θ )$] Describe and explain any patterns of equal values you see in the table for Problem 22. For Problems 25–28, + Evaluate each pair of angles to the nearest #math.equation(block: false, alt: "0.1 degrees")[$0.1^(∘)$], and show that they are supplements. + Sketch both angles. + Find the sine of each angle. #math.equation(block: true, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis")[$θ = cos^(− 1) ( display(frac(3, 4)) )$], #math.equation(block: true, alt: "ϕ equals cos to the power minus 1 ⁡ open parenthesis the fraction minus 3 over 4 close parenthesis")[$" " ϕ = cos^(− 1) ( display(frac(− 3, 4)) )$] + #math.equation(block: false, alt: "θ approximately equals 41.4 degrees , ϕ approximately equals 138.6 degrees")[$θ ≈ 41.4^(∘) , " " " " ϕ ≈ 138.6^(∘)$] + #figure(figph[angles], alt: "angles", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis ϕ close parenthesis equals the fraction the square root of 7 over 4")[$sin ( θ ) = sin ( ϕ ) = display(frac(sqrt(7), 4))$] #math.equation(block: true, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis the fraction 1 over 5 close parenthesis")[$θ = cos^(− 1) ( display(frac(1, 5)) )$], #math.equation(block: true, alt: "ϕ equals cos to the power minus 1 ⁡ open parenthesis the fraction minus 1 over 5 close parenthesis")[$" " ϕ = cos^(− 1) ( display(frac(− 1, 5)) )$] #math.equation(block: true, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis 0.1525 close parenthesis")[$θ = cos^(− 1) ( 0.1525 )$], #math.equation(block: true, alt: "ϕ equals cos to the power minus 1 ⁡ open parenthesis minus 0.1525 close parenthesis")[$" " ϕ = cos^(− 1) ( − 0.1525 )$] + #math.equation(block: false, alt: "θ approximately equals 81.2 degrees , ϕ approximately equals 98.8 degrees")[$θ ≈ 81.2^(∘) , " " " " ϕ ≈ 98.8^(∘)$] + #figure(figph[angles], alt: "angles", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis ϕ close parenthesis equals the fraction the square root of 156279 over 400 approximately equals 0.9883")[$sin ( θ ) = sin ( ϕ ) = display(frac(sqrt(156279), 400)) ≈ 0.9883$] #math.equation(block: true, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis 0.6825 close parenthesis")[$θ = cos^(− 1) ( 0.6825 )$], #math.equation(block: true, alt: "ϕ equals cos to the power minus 1 ⁡ open parenthesis minus 0.6825 close parenthesis")[$" " ϕ = cos^(− 1) ( − 0.6825 )$] For Problems 29–34, find two different angles that satisfy the equation. Round to the nearest #math.equation(block: false, alt: "0.1 degrees")[$0.1^(∘)$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.7")[$sin ( θ ) = 0.7$] #math.equation(block: false, alt: "44.4 degrees")[$44.4^(∘)$] and #math.equation(block: false, alt: "135.6 degrees")[$135.6^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.1")[$sin ( θ ) = 0.1$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over 6 equals 0.14")[$display(frac(sin ( θ ), 6)) = 0.14$] #math.equation(block: false, alt: "57.1 degrees")[$57.1^(∘)$] and #math.equation(block: false, alt: "122.9 degrees")[$122.9^(∘)$] #math.equation(block: true, alt: "the fraction 5 over sin ⁡ open parenthesis θ close parenthesis equals 6")[$display(frac(5, sin ( θ ))) = 6$] #math.equation(block: true, alt: "4.8 equals the fraction 3.2 over sin ⁡ open parenthesis θ close parenthesis")[$4.8 = display(frac(3.2, sin ( θ )))$] #math.equation(block: false, alt: "41.8 degrees")[$41.8^(∘)$] and #math.equation(block: false, alt: "138.2 degrees")[$138.2^(∘)$] #math.equation(block: true, alt: "1.5 equals the fraction sin ⁡ open parenthesis θ close parenthesis over 0.3")[$1.5 = display(frac(sin ( θ ), 0.3))$] For Problems 35–38, fill in the blanks with complements or supplements. If #math.equation(block: false, alt: "sin ⁡ open parenthesis 57 degrees close parenthesis equals q")[$" " sin ( 57^(∘) ) = q " "$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis equals q")[$" " sin ( underline(#h(3em)) ) = q " "$] also, #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals q")[$" " cos ( underline(#h(3em)) ) = q " "$], and #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals minus q")[$" " cos ( underline(#h(3em)) ) = − q$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 123 degrees close parenthesis equals q")[$sin ( 123^(∘) ) = q$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 33 degrees close parenthesis equals q")[$" " cos ( 33^(∘) ) = q$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 147 degrees close parenthesis equals minus q")[$" " cos ( 147^(∘) ) = − q$] If #math.equation(block: false, alt: "sin ⁡ open parenthesis 18 degrees close parenthesis equals w")[$" " sin ( 18^(∘) ) = w " "$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis equals w")[$" " sin ( underline(#h(3em)) ) = w " "$] also, #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals w")[$" " cos ( underline(#h(3em)) ) = w " "$], and #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals minus w")[$" " cos ( underline(#h(3em)) ) = − w$]. If #math.equation(block: false, alt: "cos ⁡ open parenthesis 74 degrees close parenthesis equals m")[$" " cos ( 74^(∘) ) = m " "$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals minus m")[$cos ( underline(#h(3em)) ) = − m " "$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis")[$" " sin ( underline(#h(3em)) ) " "$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis")[$" " sin ( underline(#h(3em)) ) " "$] both equal #math.equation(block: false, alt: "m")[$m$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis 106 degrees close parenthesis equals minus m")[$cos ( 106^(∘) ) = − m$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 16 degrees close parenthesis equals m")[$" " sin ( 16^(∘) ) = m$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 164 degrees close parenthesis equals m")[$" " sin ( 164^(∘) ) = m$] If #math.equation(block: false, alt: "cos ⁡ open parenthesis 36 degrees close parenthesis equals t")[$" " cos ( 36^(∘) ) = t " "$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals minus t")[$" " cos ( underline(#h(3em)) ) = − t " "$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis")[$" " sin ( underline(#h(3em)) )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis")[$" " sin ( underline(#h(3em)) )$] both equal #math.equation(block: false, alt: "t")[$t$]. + Sketch the line #math.equation(block: false, alt: "y equals the fraction 3 over 4 x")[$y = display(frac(3, 4)) x$]. + Find two points on the line with positive #math.equation(block: false, alt: "x")[$x$]-coordinates. + The line #math.equation(block: false, alt: "y equals the fraction 3 over 4 x")[$y = display(frac(3, 4)) x$] makes an angle with the positive #math.equation(block: false, alt: "x")[$x$]-axis. What is that angle? + Repeat parts (a) through (c) for the line #math.equation(block: false, alt: "y equals the fraction minus 3 over 4 x")[$y = display(frac(− 3, 4)) x$], except find two points with #emph[negative] #math.equation(block: false, alt: "x")[$x$]-coordinates. + #figure(figph[angles], alt: "angles", caption: none) + #math.equation(block: false, alt: "open parenthesis 4 , 3 close parenthesis , open parenthesis 8 , 6 close parenthesis")[$( 4 , 3 ) , " " ( 8 , 6 )$] + #math.equation(block: false, alt: "y equals tan to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis approximately equals 36.87 degrees")[$y = tan^(− 1) ( display(frac(3, 4)) ) ≈ 36.87^(∘)$] + #figure(figph[angles], alt: "angles", caption: none) #linebreak() #math.equation(block: false, alt: "open parenthesis minus 4 , 3 close parenthesis , open parenthesis minus 8 , 6 close parenthesis ; 143.13 degrees")[$( − 4 , 3 ) , " " ( − 8 , 6 ) ; " " 143.13^(∘)$] + Sketch the line #math.equation(block: false, alt: "y equals the fraction 5 over 3 x")[$y = display(frac(5, 3)) x$]. + Find two points on the line with positive #math.equation(block: false, alt: "x")[$x$]-coordinates. + The line #math.equation(block: false, alt: "y equals the fraction 5 over 3 x")[$y = display(frac(5, 3)) x$] makes an angle with the positive #math.equation(block: false, alt: "x")[$x$]-axis. What is that angle? + Repeat parts (a) through (c) for the line #math.equation(block: false, alt: "y equals the fraction minus 5 over 3 x")[$y = display(frac(− 5, 3)) x$], except find two points with #emph[negative] #math.equation(block: false, alt: "x")[$x$]-coordinates. For Problems 41–44, + Find exact values for the base and height of the triangle. + Compute an exact value for the area of the triangle. #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "b equals 8")[$b = 8$] in, #math.equation(block: false, alt: "h equals 3 the square root of 3")[$h = 3 sqrt(3)$] in + #math.equation(block: false, alt: "12 the square root of 3")[$12 sqrt(3)$] sq in #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "b equals 6 minus the fraction 3 the square root of 2 over 2")[$b = 6 − display(frac(3 sqrt(2), 2))$] mi, #math.equation(block: false, alt: "h equals the fraction 3 the square root of 2 over 2")[$h = display(frac(3 sqrt(2), 2))$] mi + #math.equation(block: false, alt: "the fraction 18 the square root of 2 minus 9 over 4")[$display(frac(18 sqrt(2) − 9, 4))$] sq mi #figure(figph[triangle], alt: "triangle", caption: none) Sketch an angle of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] in standard position. Find the missing coordinates of the points on the terminal side. + #math.equation(block: false, alt: "open parenthesis minus 1 , ? close parenthesis")[$( − 1 , ? )$] + #math.equation(block: false, alt: "open parenthesis ? , 3 close parenthesis")[$( ? , 3 )$] #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 1 , the square root of 3 close parenthesis")[$( − 1 , sqrt(3) )$] + #math.equation(block: false, alt: "open parenthesis minus the square root of 3 , 3 close parenthesis")[$( − sqrt(3) , 3 )$] Sketch an angle of #math.equation(block: false, alt: "150 degrees")[$150^(∘)$] in standard position. Find the missing coordinates of the points on the terminal side. + #math.equation(block: false, alt: "open parenthesis ? , 2 close parenthesis")[$( ? , 2 )$] + #math.equation(block: false, alt: "open parenthesis minus 4 , ? close parenthesis")[$( − 4 , ? )$] Sketch an angle of #math.equation(block: false, alt: "135 degrees")[$135^(∘)$] in standard position. Find the missing coordinates of the points on the terminal side. + #math.equation(block: false, alt: "open parenthesis ? , 3 close parenthesis")[$( ? , 3 )$] + #math.equation(block: false, alt: "open parenthesis minus the square root of 5 , ? close parenthesis")[$( − sqrt(5) , ? )$] #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 3 , 3 close parenthesis")[$( − 3 , 3 )$] + #math.equation(block: false, alt: "open parenthesis minus the square root of 5 , the square root of 5 close parenthesis")[$( − sqrt(5) , sqrt(5) )$] + Use a sketch to explain why #math.equation(block: false, alt: "cos ⁡ open parenthesis 90 degrees close parenthesis equals 0")[$cos ( 90^(∘) ) = 0$]. + Use a sketch to explain why #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees close parenthesis equals 1")[$cos ( 180^(∘) ) = 1$]. For Problems 49–54, find the area of the triangle with the given properties. Round your answer to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: false, alt: "20.71")[$20.71$] sq m #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: false, alt: "55.51")[$55.51$] sq cm #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: false, alt: "a equals 0.8")[$a = 0.8$] m, #math.equation(block: false, alt: "c equals 0.15")[$c = 0.15$] m, #math.equation(block: false, alt: "B equals 15 degrees")[$B = 15^(∘)$] Find the area of the regular pentagon shown at right. (Hint: The pentagon can be divided into five congruent triangles.) #figure(figph[pentagon], alt: "pentagon", caption: none) #math.equation(block: false, alt: "38.04")[$38.04$] sq units Find the area of the regular hexagon shown at right. (Hint: The hexagon can be divided into six congruent triangles.) #figure(figph[hexagon], alt: "hexagon", caption: none) For Problems 57 and 58, lots from a housing development have been subdivided into triangles. Find the total area of each lot by computing and adding the areas of each triangle. #figure(figph[lot], alt: "lot", caption: none) #math.equation(block: false, alt: "13 , 851.3")[$13 , 851.3$] sq ft #figure(figph[lot], alt: "lot", caption: none) For Problems 59 and 60, + Find the coordinates of point #math.equation(block: false, alt: "P")[$P$]. Round to two decimal places. + Find the sides #math.equation(block: false, alt: "B C")[$B C$] and #math.equation(block: false, alt: "P C")[$P C$] of #math.equation(block: false, alt: "△ P C B")[$upright(△) P C B$]. + Find side #math.equation(block: false, alt: "P B")[$P B$]. #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 74.97 , 59.00 close parenthesis")[$( − 74.97 , 59.00 )$] + #math.equation(block: false, alt: "B C equals 141.97 , P C equals 59.00")[$B C = 141.97 , " " " " P C = 59.00$] + #math.equation(block: false, alt: "153.74")[$153.74$] #figure(figph[triangle], alt: "triangle", caption: none) Later we will be able to show that #math.equation(block: false, alt: "sin ⁡ open parenthesis 18 degrees close parenthesis equals the fraction the square root of 5 minus 1 over 4")[$sin ( 18^(∘) ) = display(frac(sqrt(5) − 1, 4))$]. What is the exact value of #math.equation(block: false, alt: "sin ⁡ open parenthesis 162 degrees close parenthesis ?")[$sin ( 162^(∘) ) ?$] (Hint: Sketch both angles in standard position.) #math.equation(block: true, alt: "the fraction the square root of 5 minus 1 over 4")[$display(frac(sqrt(5) − 1, 4))$] Later we will be able to show that #math.equation(block: false, alt: "cos ⁡ open parenthesis 36 degrees close parenthesis equals the fraction the square root of 5 plus 1 over 4")[$cos ( 36^(∘) ) = display(frac(sqrt(5) + 1, 4))$]. What is the exact value of #math.equation(block: false, alt: "cos ⁡ open parenthesis 144 degrees close parenthesis ?")[$cos ( 144^(∘) ) ?$] (Hint: Sketch both angles in standard position.) Alice wants an obtuse angle #math.equation(block: false, alt: "θ")[$θ$] that satisfies #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.3")[$sin ( θ ) = 0.3$]. Bob presses some buttons on his calculator and reports that #math.equation(block: false, alt: "θ equals 17.46 degrees")[$θ = 17.46^(∘)$]. Explain Bob's error and give a correct approximation of #math.equation(block: false, alt: "θ")[$θ$] accurate to two decimal places. Bob found an acute angle. The obtuse angle is the supplement of #math.equation(block: false, alt: "17.46 degrees")[$17.46^(∘)$], or #math.equation(block: false, alt: "162.54 degrees")[$162.54^(∘)$]. Yaneli finds that the angle #math.equation(block: false, alt: "θ")[$θ$] opposite the longest side of a triangle satisfies #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.8")[$sin ( θ ) = 0.8$]. Zelda reports that #math.equation(block: false, alt: "θ equals 53.13 degrees")[$θ = 53.13^(∘)$]. Explain Zelda's error and give a correct approximation of #math.equation(block: false, alt: "θ")[$θ$] accurate to two decimal places. For Problems 65–70, + Sketch an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position, #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 180 degrees")[$0^(∘) ≤ θ ≤ 180^(∘)$], with the given properties. + Find expressions for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis")[$cos ( θ ) , " " sin ( θ )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] in terms of the given variable. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over 3 , x less than 0")[$cos ( θ ) = display(frac(x, 3)) , " " x < 0$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over 3")[$cos ( θ ) = display(frac(x, 3))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 9 minus x squared over 3")[$" " sin ( θ ) = display(frac(sqrt(9 − x^(2)), 3))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 9 minus x squared over x")[$" " tan ( θ ) = display(frac(sqrt(9 − x^(2)), x))$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 4 over α , α less than 0")[$tan ( θ ) = display(frac(4, α)) , " " α < 0$] #math.equation(block: false, alt: "θ")[$θ$] is obtuse and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction y over 2")[$sin ( θ ) = display(frac(y, 2))$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 4 minus y squared over 2")[$cos ( θ ) = display(frac(− sqrt(4 − y^(2)), 2))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction y over 2")[$" " sin ( θ ) = display(frac(y, 2))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus y over the square root of 4 minus y squared")[$" " tan ( θ ) = display(frac(− y, sqrt(4 − y^(2))))$] #math.equation(block: false, alt: "θ")[$θ$] is obtuse and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction q over minus 7")[$tan ( θ ) = display(frac(q, − 7))$] #math.equation(block: false, alt: "θ")[$θ$] is obtuse and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals m")[$tan ( θ ) = m$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over the square root of 1 plus m squared")[$cos ( θ ) = display(frac(− 1, sqrt(1 + m^(2))))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus m over the square root of 1 plus m squared")[$" " sin ( θ ) = display(frac(− m, sqrt(1 + m^(2))))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals m")[$" " tan ( θ ) = m$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals h")[$cos ( θ ) = h$]