#set document(title: "2.2 Right Triangle Trigonometry", 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")) == 2.2#h(0.6em)Right Triangle Trigonometry With the Pythagorean theorem we can find one side of a right triangle if we know the other #emph[two] sides. By using what we know about similar triangles, we can find the unknown sides of a right triangle if we know only #emph[one] side and #emph[one] of the acute angles. #figure(figph[right triangle with one unknown side], alt: "right triangle with one unknown side", caption: none) #figure(figph[right triangle with two unknown sides], alt: "right triangle with two unknown sides", caption: none) #math.equation(block: true, alt: "We can find side b with the Pythagorean theorem.")[$"We can find side" " " bold(italic(b)) " " "with the Pythagorean theorem."$] #math.equation(block: true, alt: "Can we find side b ?")[$"Can we find side" " " bold(italic(b)) "?"$] === The Sine of an Angle In Example 2 of Section 1.2, we saw that in a 30-60-90 right triangle, the ratio of the shortest side to the hypotenuse was #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$], or 0.5. This ratio is the same for any two right triangles with a #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angle, because they are similar triangles, as shown at right. #figure(figph[30-60-90 right triangle], alt: "30-60-90 right triangle", caption: none) The ratio is given a name; it is called the "sine of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]." We write #math.equation(block: true, alt: "sin ⁡ open parenthesis 30 degrees close parenthesis equals 0.5 ,")[$sin ( 30^(∘) ) = 0.5 ,$] where #emph[sin] is an abbreviation for sine. There is nothing special about #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angles; we can talk about the sine of any angle. The #strong[sine] of an angle is the ratio of the side opposite the angle to the hypotenuse. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Recall that we use the notation #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] to indicate a #strong[function] of an input variable, #math.equation(block: false, alt: "x")[$x$]. In fact, the sine ratio is a function of the angle, so we use parentheses around the angle input to refer to its sine, like this: #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]. ] #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sine of an Acute Angle] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction opposite over hypotenuse")[$" " " " " " " " " " " " sin ( θ ) = display(frac("opposite", "hypotenuse"))$] #figure(figph[right triangle], alt: "right triangle", caption: none) ] #examplebox("Example 1")[][ Find the sine of the labeled angle in each triangle below. #figure(figph[right triangle], alt: "right triangle", caption: none) #figure(figph[right triangle], alt: "right triangle", caption: none) #solutionbox[ + The side opposite angle #math.equation(block: false, alt: "ϕ")[$ϕ$] is 5, and the hypotenuse is 13, so the sine of #math.equation(block: false, alt: "ϕ")[$ϕ$] is #math.equation(block: true, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 5 over 13 , or approximately 0.3846")[$sin ( ϕ ) = display(frac(5, 13)) , " or approximately" " " " " 0.3846$] + The side opposite angle #math.equation(block: false, alt: "β")[$β$] is #math.equation(block: false, alt: "the square root of 5")[$sqrt(5)$], and the hypotenuse is 3, so the sine of #math.equation(block: false, alt: "β")[$β$] is #math.equation(block: true, alt: "sin ⁡ open parenthesis β close parenthesis equals the fraction the square root of 5 over 3 , or approximately 0.7454")[$sin ( β ) = display(frac(sqrt(5), 3)) , " or approximately" " " " " 0.7454$] ] ] Find the sine of the labeled angle in the triangle at right. Round your answer to 4 decimal places. #figure(figph[right triangle], alt: "right triangle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction 6.4 over 8.6 approximately equals 0.7442")[$sin ( α ) = display(frac(6.4, 8.6)) ≈ 0.7442$] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We must use the sides of a #emph[right] triangle to calculate the sine of an angle. For example, in the triangle at right, #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 7")[$sin ( θ ) = frac(4, 7)$], because #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] is a right triangle. It is #emph[not] true that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 7")[$sin ( θ ) = frac(5, 7)$], or #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 6 over 7")[$sin ( θ ) = frac(6, 7)$]. In this chapter, we consider only right triangles. #figure(figph[triangles], alt: "triangles", caption: none) ] === Using a Calculator Mathematicians have calculated the sines of any angle we like. The values of the sine were originally collected into tables, and are now available on scientific calculators. For example, let's find the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]. First, consider some triangles, as shown below. #figure(figph[right triangles], alt: "right triangles", caption: none) #math.equation(block: true, alt: "Which angle has the larger sine, 30 degrees or 50 degrees ?")[$"Which angle has the larger sine," " " 30^(∘) " " "or" " " 50^(∘) ?$] Do you expect the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] to be larger or smaller than the sine of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]? Do you expect the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] to be larger or smaller than 1? #examplebox("Example 2")[][ Use your calculator to find the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] by entering SIN #math.equation(block: false, alt: "open parenthesis 50 close parenthesis")[$( 50 )$]. (Make sure your calculator is set for degrees.) You should find that #math.equation(block: true, alt: "sin ⁡ open parenthesis 50 degrees close parenthesis equals 0.7660444431")[$sin ( 50^(∘) ) = 0.7660444431$] This is not an exact value; the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] is an irrational number, and your calculator shows as many digits as its display will allow. (Not all sine values are as “nice” as the sine of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]!) Usually we round to four decimal places, so we write #math.equation(block: true, alt: "sin ⁡ open parenthesis 50 degrees close parenthesis equals 0.7660")[$sin ( 50^(∘) ) = 0.7660$] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Note that when you press the sine key, SIN, your calculator displays #math.equation(block: true, alt: "sin open parenthesis")[$"sin" " " \($] with an open parenthesis, as the prompt to enter an angle, because the sine is a function of the angle. You can think of the notation #emph[sin] as an operation symbol telling you to find the sine of an angle, just as the symbol #math.equation(block: false, alt: "the square root of")[$sqrt(" " " ")$] tells you to take the square root of the expression under the radical. ] + Use your calculator to complete the table, rounding your answers to four decimal places.#figure(table( columns: 11, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0 degrees")[$" " " " 0^(∘)$]], [#math.equation(block: false, alt: "10 degrees")[$" " 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^(∘)$]], [#math.equation(block: false, alt: "90 degrees")[$" " 90^(∘)$]]), [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], )) + What do you notice about the values of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] to #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]? If you plot the values of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] against the values of #math.equation(block: false, alt: "θ")[$θ$], will the graph be a straight line? Why or why not? + #figure(table( columns: 11, 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: "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: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]]), [#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: "0.1737")[$0.1737$]], [#math.equation(block: false, alt: "0.3420")[$0.3420$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.6428")[$0.6428$]], [#math.equation(block: false, alt: "0.7660")[$0.7660$]], [#math.equation(block: false, alt: "0.8660")[$0.8660$]], [#math.equation(block: false, alt: "0.9397")[$0.9397$]], [#math.equation(block: false, alt: "0.9848")[$0.9848$]], [#math.equation(block: false, alt: "1")[$" " " " 1$]], )) + The values of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] increase from 0 to 1 as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] to #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. The graph will not be a straight line because the slopes between successive points are not constant. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The important thing to remember is that the sine of an angle, say #math.equation(block: false, alt: "50 degrees")[$50^(∘)$], is the same for any right triangle with a #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] angle, no matter what the size or orientation of the triangle. The figure below shows three different right triangles with a #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] angle. Although the sides of the triangle may be bigger or smaller, the #emph[ratio] #math.equation(block: false, alt: "the fraction opposite over hypotenuse")[$display(frac("opposite", "hypotenuse"))$] is always the same for that angle, because the triangles are similar. This is why the sine ratio is useful. #figure(figph[right triangles, 50 degrees], alt: "right triangles, 50 degrees", caption: none) #math.equation(block: true, alt: "In each triangle, the ratio sin ⁡ open parenthesis 50 degrees close parenthesis equals the fraction opposite over hypotenuse equals 0.7660")[$"In each triangle, the ratio" " " bold(sin) ( 50^(∘) ) = display(frac("opposite", "hypotenuse")) = 0.7660$] ] === Using the Sine Ratio to Find an Unknown Side In the next example we see how to use the sine ratio to find an unknown side in a right triangle, knowing only one other side and one angle. #examplebox("Example 3")[][ Find the length of the side opposite the #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] angle in the triangle shown. #figure(figph[right triangle, 50 degrees], alt: "right triangle, 50 degrees", caption: none) #solutionbox[ In this triangle, the ratio #math.equation(block: false, alt: "the fraction opposite over hypotenuse")[$display(frac("opposite", "hypotenuse"))$] is equal to the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$], or #math.equation(block: true, alt: "sin ⁡ open parenthesis 50 degrees close parenthesis equals the fraction opposite over hypotenuse")[$sin ( 50^(∘) ) = display(frac("opposite", "hypotenuse"))$] We use a calculator to find an approximate value for the sine of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$], filling in the lengths of the hypotenuse and the opposite side to get #math.equation(block: true, alt: "0.7660 equals the fraction x over 18")[$0.7660 = display(frac(x, 18))$] We solve for #math.equation(block: false, alt: "x")[$x$] to find #math.equation(block: true, alt: "x equals 18 open parenthesis 0.7660 close parenthesis equals 13.788")[$x = 18 ( 0.7660 ) = 13.788$] To two decimal places, the length of the opposite side is 13.79 centimeters. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the previous example, even though we showed only four places in #math.equation(block: false, alt: "sin ⁡ open parenthesis 50 degrees close parenthesis")[$sin ( 50^(∘) )$], you should not round off intermediate steps in a calculation, because the answer loses accuracy with each rounding. You can use the following keystrokes on your calculator to avoid entering a long approximation for #math.equation(block: false, alt: "sin ⁡ open parenthesis 50 degrees close parenthesis")[$sin ( 50^(∘) )$]: #math.equation(block: true, alt: "sin open parenthesis 50 close parenthesis times 18")[$"sin" " " ( 50 ) " " × 18$] The calculator returns #math.equation(block: false, alt: "x equals 13.78879998.")[$" " x = 13.78879998 .$] ] Find the length of the hypotenuse in the triangle shown. #figure(figph[right triangle, 62 degrees], alt: "right triangle, 62 degrees", caption: none) #math.equation(block: false, alt: "sin ⁡ open parenthesis 62 degrees close parenthesis equals the fraction 7.5 over x")[$sin ( 62^(∘) ) = display(frac(7.5, x))$]. Solve this equation to find #math.equation(block: false, alt: "x equals the fraction 7.5 over 0.8829 approximately equals 8.49")[$" " x = display(frac(7.5, 0.8829)) ≈ 8.49$], or, rounding to one decimal place, about 8.5 m === The Cosine and the Tangent There are two more trigonometric ratios used for calculating the sides of right triangles, depending on which of the three sides is known and which are unknown. These ratios are called the #strong[cosine] and the #strong[tangent]. Suppose we'd like to find the height of a tall cliff without actually climbing it. We can measure the distance to the base of the cliff, and we can use a surveying tool called a theodolite to measure the angle between the ground and our line of sight to the top of the cliff (this is called the #strong[angle of elevation]). These values give us two parts of a right triangle, as shown at right. The height we want is the side opposite the angle of elevation. The distance to the base of the cliff is the length of the side #strong[adjacent] to the angle of elevation. #figure(figph[angle of elevation triangle], alt: "angle of elevation triangle", caption: none) The ratio of the side opposite an angle to the side adjacent to the angle is called the #strong[tangent] of the angle. The abbreviation for “tangent of theta” is tan (#math.equation(block: false, alt: "θ")[$θ$]). #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Tangent of an Acute Angle] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction opposite over adjacent")[$#h(2em) tan ( θ ) = display(frac("opposite", "adjacent"))$] #figure(figph[right triangle], alt: "right triangle", caption: none) ] Just like the sine of an angle, the tangent ratio is always the same for a given angle, no matter what size triangle it occurs in. And just like #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$], we can find the values of #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] on a scientific calculator. #examplebox("Example 4")[][ + Use your calculator to find the tangent of #math.equation(block: false, alt: "58 degrees")[$58^(∘)$]. + Find the height of the cliff if the angle of elevation to the top of the cliff is #math.equation(block: false, alt: "58 degrees")[$58^(∘)$] at a distance of 300 feet from the base of the cliff. #solutionbox[ + Enter TAN (58) to find #math.equation(block: false, alt: "tan ⁡ open parenthesis 58 degrees close parenthesis equals 1.6003")[$tan ( 58^(∘) ) = 1.6003$], rounded to four decimal places. + We use the tangent ratio to write an equation. In this triangle, the angle is #math.equation(block: false, alt: "58 degrees")[$58^(∘)$]. #math.equation(block: true, alt: "tan ⁡ open parenthesis 58 degrees close parenthesis equals the fraction opposite over adjacent")[$tan ( 58^(∘) ) = display(frac("opposite", "adjacent"))$]#figure(figph[angle of elevation triangle], alt: "angle of elevation triangle", caption: none) #linebreak() Next we fill in the value of #math.equation(block: false, alt: "tan ⁡ open parenthesis 58 degrees close parenthesis")[$tan ( 58^(∘) )$], and the lengths of the sides. #math.equation(block: true, alt: "1.6003 equals the fraction h over 300")[$1.6003 = frac(h, 300)$] Solving for #math.equation(block: false, alt: "h")[$h$] gives #math.equation(block: true, alt: "h equals 300 open parenthesis 1 , 6003 close parenthesis equals 480.0")[$h = 300 ( 1 , 6003 ) = 480.0$] so the height of the cliff is about 480 feet. ] ] + Use the tangent ratio to find #math.equation(block: false, alt: "x")[$x$] in the triangle shown. #figure(figph[right triangle], alt: "right triangle", caption: none) + Use the sine ratio to find the hypotenuse, #math.equation(block: false, alt: "c")[$c$], of the triangle. + Use the Pythagorean theorem to find the hypotenuse of the triangle. Do you get the same answer with both methods? Can you explain why the calculations might give (slightly) different answers? + #math.equation(block: false, alt: "tan ⁡ open parenthesis 24.7 degrees close parenthesis equals the fraction x over 50")[$tan ( 24.7^(∘) ) = display(frac(x, 50))$], and solving this equation gives #math.equation(block: false, alt: "x equals 23")[$x = 23$] ft. + #math.equation(block: false, alt: "sin ⁡ open parenthesis 24.7 degrees close parenthesis equals the fraction 23 over c")[$sin ( 24.7^(∘) ) = display(frac(23, c))$], so #math.equation(block: false, alt: "c equals 55")[$c = 55$] ft. + The answers agree when rounded to units. Rounding during calculation can cause the results to differ. The third trigonometric ratio, called the #strong[cosine], is the ratio of the side adjacent to an angle and the hypotenuse of the triangle. #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Cosine of an Acute Angle] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction adjacent over hypotenuse")[$#h(2em) cos ( θ ) = display(frac("adjacent", "hypotenuse"))$] #figure(figph[right triangle], alt: "right triangle", caption: none) ] #examplebox("Example 5")[][ Find #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")[$tan ( θ )$] for the triangle shown at right. #figure(figph[right triangle], alt: "right triangle", caption: none) #solutionbox[ First, we use the Pythagorean theorem to find the hypotenuse, #math.equation(block: false, alt: "c")[$c$]. #math.equation(block: true, alt: "c squared, equals 6 squared plus 8 squared; c squared, equals 536 equals 84 equals 100, Take square roots.; c, equals the square root of 100 equals 10")[$c^(2) & = 6^(2) + 8^(2) \ c^(2) & = 536 = 84 = 100 & & "Take square roots." \ c & = sqrt(100) = 10$] For the angle #math.equation(block: false, alt: "θ")[$θ$], the opposite side is 8 inches long, and the adjacent side is 6 inches long, as shown in the figure. Thus, #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis, equals the fraction opposite over hypotenuse equals the fraction 8 over 10 or 0.8; cos ⁡ open parenthesis θ close parenthesis, equals the fraction adjacent over hypotenuse equals the fraction 6 over 10 or 0.6; tan ⁡ open parenthesis θ close parenthesis, equals the fraction opposite over adjacent equals the fraction 8 over 6 or 1.3 bar")[$sin ( θ ) & = display(frac("opposite", "hypotenuse")) = frac(8, 10) " " " " "or" " " " " 0.8 \ cos ( θ ) & = display(frac("adjacent", "hypotenuse")) = frac(6, 10) " " " " "or" " " " " 0.6 \ tan ( θ ) & = display(frac("opposite", "adjacent")) = frac(8, 6) " " " " "or" " " " " 1. limits(3)^(―)$] ] ] #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sine and Cosine] + Use your calculator to complete the table. Round the values of sine and cosine to four decimal places.#figure(table( columns: 11, 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: "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")[$" " 80^(∘)$]], [#math.equation(block: false, alt: "90 degrees")[$" " 90^(∘)$]]), [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], )) + What do you notice about the values of sine and cosine? Can you explain why this is true? (Hint: If one (non-right) angle of a right triangle measures #math.equation(block: false, alt: "x")[$x$] degrees, how big is the other angle?) + Sketch a right triangle and label the opposite and adjacent sides for each (non-right) angle. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the Activity, you should also notice that as the angle #math.equation(block: false, alt: "θ")[$θ$] increases, #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] increases but #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] decreases. You can see why this is true in the figure below. In each right triangle, the hypotenuse has the same length. But as the angle increases, the opposite side gets longer and the adjacent side gets shorter. The figure this passage describes, with the angle in your hands and the hypotenuse pinned at length 1. The blue hypotenuse reaches from the origin to the dashed circular arc of radius 1, so its length never changes; the red horizontal leg is the side adjacent to the angle θ, and the red vertical leg is the side opposite. Because the hypotenuse is 1, the ratio opposite/hypotenuse is just the opposite leg's own length — the legs ARE sin(θ) and cos(θ). Drag θ from 10° up to 80°: the opposite leg grows (the sine increases) while the adjacent leg shrinks (the cosine decreases), which is exactly the observation in the text. At θ = 30° the opposite leg is exactly half the hypotenuse — sin(30°) = 0.5, the ratio the section computed from the 30-60-90 triangle.{"functions":\[{"color":"\#6a6a6a","expression":"cos(t)","expression2":"sin(t)","id":"unit-arc","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":1.5708,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*cos(a\*pi/180)","expression2":"t\*sin(a\*pi/180)","id":"hypotenuse","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"t\*cos(a\*pi/180)","expression2":"0","id":"adjacent-leg","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"cos(a\*pi/180)","expression2":"t\*sin(a\*pi/180)","id":"opposite-leg","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#2563eb","expression":"0.18\*cos(a\*pi/180\*t)","expression2":"0.18\*sin(a\*pi/180\*t)","id":"angle-arc","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"cos(a\*pi/180) + 0.04\*cos(t)","expression2":"sin(a\*pi/180) + 0.04\*sin(t)","id":"tip","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Angle \\u03b8","max":80,"min":10,"name":"a","step":5,"unit":"\\u00b0","value":30}\],"title":"Right triangle with hypotenuse 1: legs cos(\\u03b8) and sin(\\u03b8)","version":1,"viewport":{"centerX":0.45,"centerY":0.4,"scale":220}}#figure(figph[right triangles with increasing theta], alt: "right triangles with increasing theta", caption: none) ] === The Three Trigonometric Ratios Here is a summary of the three trigonometric ratios we have discussed. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Trigonometric Ratios] If #math.equation(block: false, alt: "θ")[$θ$] is one of the angles in a right triangle, #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis, equals the fraction opposite over hypotenuse; cos ⁡ open parenthesis θ close parenthesis, equals the fraction adjacent over hypotenuse; tan ⁡ open parenthesis θ close parenthesis, equals the fraction opposite over adjacent")[$sin ( θ ) & = display(frac("opposite", "hypotenuse")) \ cos ( θ ) & = display(frac("adjacent", "hypotenuse")) \ tan ( θ ) & = display(frac("opposite", "adjacent"))$] #figure(figph[right triangle], alt: "right triangle", caption: none) ] These three definitions are the foundation for all the rest of trigonometry. #strong[You should internalize them immediately!!] We must also be careful to apply these definitions of the trigonometric ratios #emph[only] to right triangles. In the next example, we create a right triangle by drawing an extra line. #examplebox("Example 6")[][ The vertex angle of an isosceles triangle is #math.equation(block: false, alt: "34 degrees")[$34^(∘)$], and the equal sides are 16 meters long. Find the altitude of the triangle. #solutionbox[ The triangle described is not a right triangle. However, the altitude of an isosceles triangle bisects the vertex angle and divides the triangle into two congruent right triangles, as shown in the figure. The 16-meter side becomes the hypotenuse of the right triangle, and the altitude, #math.equation(block: false, alt: "h")[$h$], of original triangle is the side adjacent to the #math.equation(block: false, alt: "17 degrees")[$17^(∘)$] angle. #figure(figph[triangle with altitude], alt: "triangle with altitude", caption: none) Which of the three trig ratios is helpful in this problem? The cosine is the ratio that relates the hypotenuse and the adjacent side, so we'll begin with the equation #math.equation(block: true, alt: "cos ⁡ open parenthesis 17 degrees close parenthesis equals the fraction adjacent over hypotenuse")[$cos ( 17^(∘) ) = display(frac("adjacent", "hypotenuse"))$] We use a calculator to find #math.equation(block: false, alt: "cos ⁡ open parenthesis 17 degrees close parenthesis")[$cos ( 17^(∘) )$] and fill in the lengths of the sides. #math.equation(block: true, alt: "0.9563 equals the fraction h over 16")[$0.9563 = display(frac(h, 16))$] Solving for #math.equation(block: false, alt: "h")[$h$] gives #math.equation(block: true, alt: "h equals 16 open parenthesis 0.9563 close parenthesis equals 15.3008")[$h = 16 ( 0.9563 ) = 15.3008$] The altitude of the triangle is about 15.3 meters long. ] ] Another isosceles triangle has base angles of #math.equation(block: false, alt: "72 degrees")[$72^(∘)$] and equal sides of length 6.8 centimeters. Find the length of the base. Let #math.equation(block: false, alt: "x")[$x$] stand for half of the base. Then #math.equation(block: false, alt: "cos ⁡ open parenthesis 72 degrees close parenthesis equals the fraction x over 6.8")[$cos ( 72^(∘) ) = display(frac(x, 6.8))$], so #math.equation(block: false, alt: "x equals 2.1")[$x = 2.1$] and the base is 4.2 cm. Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write two more ratios equivalent to the given fraction. 1. #math.equation(block: false, alt: "the fraction 10 over 4")[$#h(0.222em) display(frac(10, 4))$] 2. #math.equation(block: false, alt: "the fraction 6 over 8")[$#h(0.222em) display(frac(6, 8))$] 3. #math.equation(block: false, alt: "0.6")[$#h(0.222em) 0.6$] 4. #math.equation(block: false, alt: "1.5")[$#h(0.222em) 1.5$] Compute the slope of the line. 5. #figure(figph[grid], alt: "grid", caption: none) 6. #figure(figph[grid], alt: "grid", caption: none) 7. #figure(figph[grid], alt: "grid", caption: none) 8. #figure(figph[grid], alt: "grid", caption: none) Solve. 9. #math.equation(block: false, alt: "the fraction 12 over x equals 48")[$#h(0.222em) display(frac(12, x)) = 48$] 10. #math.equation(block: false, alt: "the fraction 60 over x equals 80")[$#h(0.222em) display(frac(60, x)) = 80$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers (Many answers are possible for 1–4.) + #math.equation(block: false, alt: "the fraction 5 over 2 , the fraction 20 over 8")[$display(frac(5, 2)) , " " " " display(frac(20, 8))$] + #math.equation(block: false, alt: "the fraction 3 over 4 , the fraction 12 over 16")[$display(frac(3, 4)) , " " " " display(frac(12, 16))$] + #math.equation(block: false, alt: "the fraction 3 over 5 , the fraction 12 over 20")[$display(frac(3, 5)) , " " " " display(frac(12, 20))$] + #math.equation(block: false, alt: "the fraction 6 over 4 , the fraction 12 over 8")[$display(frac(6, 4)) , " " " " display(frac(12, 8))$] + #math.equation(block: false, alt: "the fraction 2 over 5")[$display(frac(2, 5))$] + #math.equation(block: false, alt: "the fraction minus 8 over 5")[$display(frac(− 8, 5))$] + #math.equation(block: false, alt: "the fraction minus 6 over 5")[$display(frac(− 6, 5))$] + #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] + #math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$] + #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] ] === Section 2.2 Summary ==== Vocabulary - Sine - Cosine - Tangent - Angle of elevation - Adjacent side - Irrational number ==== Concepts + By using similar triangles, we can find the unknown sides of a right triangle if we know only #emph[one] side and #emph[one] of the acute angles. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Trigonometric Ratios] If #math.equation(block: false, alt: "θ")[$θ$] is one of the angles in a right triangle, #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis, equals the fraction opposite over hypotenuse; cos ⁡ open parenthesis θ close parenthesis, equals the fraction adjacent over hypotenuse; tan ⁡ open parenthesis θ close parenthesis, equals the fraction opposite over adjacent")[$sin ( θ ) & = display(frac("opposite", "hypotenuse")) \ cos ( θ ) & = display(frac("adjacent", "hypotenuse")) \ tan ( θ ) & = display(frac("opposite", "adjacent"))$]#figure(figph[triangle], alt: "triangle", caption: none) ] + The trigonometric ratio of an angle #math.equation(block: false, alt: "θ")[$θ$] is the same for every right triangle containing the angle. ==== Study Questions + Sketch a figure that illustrates why #math.equation(block: false, alt: "cos ⁡ open parenthesis 25 degrees close parenthesis")[$cos ( 25^(∘) )$] is the same for every right triangle with a #math.equation(block: false, alt: "25 degrees")[$25^(∘)$] angle. + Sketch a figure that illustrates why #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] decreases as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] to #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. + Which trigonometric ratio would you use to find the hypotenuse of a right triangle if you knew one acute angle and the side opposite that angle? + Does your calculator give you the exact decimal values for the trigonometric ratios of acute angles? ==== Skills Practice each skill in the Homework Problems listed. + Use measurements to calculate the trigonometric ratios for acute angles \#1-10, 57-60 + Use trigonometric ratios to find unknown sides of right triangles \#11-26 + Solve problems using trigonometric ratios \#27-34, 41-46 + Use trig ratios to write equations relating the sides of a right triangle \#35-40 + Use relationships among the trigonometric ratios \#47-56, 61-68 === Homework 2.2 Here are two right triangles with a #math.equation(block: false, alt: "65 degrees")[$65^(∘)$] angle. + Measure the sides #math.equation(block: false, alt: "A B")[$A B$] and #math.equation(block: false, alt: "B C")[$B C$] with a ruler. Use the lengths to estimate #math.equation(block: false, alt: "sin ⁡ open parenthesis 65 degrees close parenthesis")[$sin ( 65^(∘) )$]. + Measure the sides #math.equation(block: false, alt: "A D")[$A D$] and #math.equation(block: false, alt: "D E")[$D E$] with a ruler. Use the lengths to estimate #math.equation(block: false, alt: "sin ⁡ open parenthesis 65 degrees close parenthesis")[$sin ( 65^(∘) )$]. + Use your calculator to look up #math.equation(block: false, alt: "sin ⁡ open parenthesis 65 degrees close parenthesis")[$sin ( 65^(∘) )$]. Compare your answers. How close were your estimates? #figure(figph[triangles], alt: "triangles", caption: none) + 0.91 + 0.91 + 0.9063 Use the figure in Problem 1 to calculate two estimates each for the cosine and tangent of #math.equation(block: false, alt: "65 degrees")[$65^(∘)$]. Compare your estimates to your calculator's values for #math.equation(block: false, alt: "cos ⁡ open parenthesis 65 degrees close parenthesis")[$cos ( 65^(∘) )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 65 degrees close parenthesis")[$tan ( 65^(∘) )$]. Here are two right triangles with a #math.equation(block: false, alt: "40 degrees")[$40^(∘)$] angle. + Measure the sides #math.equation(block: false, alt: "A B")[$A B$] and #math.equation(block: false, alt: "A C")[$A C$] with a ruler. Use the lengths to estimate #math.equation(block: false, alt: "cos ⁡ open parenthesis 40 degrees close parenthesis")[$cos ( 40^(∘) )$]. + Measure the sides #math.equation(block: false, alt: "A D")[$A D$] and #math.equation(block: false, alt: "A E")[$A E$] with a ruler. Use the lengths to estimate #math.equation(block: false, alt: "cos ⁡ open parenthesis 40 degrees close parenthesis")[$cos ( 40^(∘) )$]. + Use your calculator to look up #math.equation(block: false, alt: "cos ⁡ open parenthesis 40 degrees close parenthesis")[$cos ( 40^(∘) )$]. Compare your answers. How close were your estimates? #figure(figph[triangles], alt: "triangles", caption: none) + 0.77 + 0.77 + 0.7660 Use the figure in Problem 2 to calculate two estimates each for the cosine and tangent of #math.equation(block: false, alt: "40 degrees")[$40^(∘)$]. Compare your estimates to your calculator's values for #math.equation(block: false, alt: "sin ⁡ open parenthesis 40 degrees close parenthesis")[$sin ( 40^(∘) )$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis 40 degrees close parenthesis")[$tan ( 40^(∘) )$]. For the right triangles in Problems 5–10, + Find the length of the unknown side. + Find the sine, cosine, and tangent of #math.equation(block: false, alt: "θ")[$θ$]. Round your answers to four decimal places. #figure(figph[right triangle], alt: "right triangle", caption: none) \> + #math.equation(block: false, alt: "4 the square root of 13 approximately equals 14.42")[$4 sqrt(13) ≈ 14.42$] + #math.equation(block: false, alt: "sin ⁡ θ equals 0.5547")[$sin θ = 0.5547$], #math.equation(block: false, alt: "cos ⁡ θ equals 0.8321")[$cos θ = 0.8321$], #math.equation(block: false, alt: "tan ⁡ θ equals 0.6667")[$tan θ = 0.6667$] #figure(figph[right triangle], alt: "right triangle", caption: none) #figure(figph[right triangle], alt: "right triangle", caption: none) + #math.equation(block: false, alt: "4 the square root of 15 approximately equals 15.49")[$4 sqrt(15) ≈ 15.49$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.9682")[$sin ( θ ) = 0.9682$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.2500")[$cos ( θ ) = 0.2500$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals 3.8730")[$tan ( θ ) = 3.8730$] #figure(figph[right triangle], alt: "right triangle", caption: none) #figure(figph[right triangle], alt: "right triangle", caption: none) + #math.equation(block: false, alt: "2 the square root of 67 approximately equals 16.37")[$2 sqrt(67) ≈ 16.37$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.2116")[$sin ( θ ) = 0.2116$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.9774")[$cos ( θ ) = 0.9774$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals 0.2165")[$tan ( θ ) = 0.2165$] #figure(figph[right triangle], alt: "right triangle", caption: none) For Problems 11–16, + Sketch and label the sides of a right triangle with angle #math.equation(block: false, alt: "θ")[$θ$]. + Sketch and label another right triangle with angle #math.equation(block: false, alt: "θ")[$θ$] and longer sides. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5")[$cos ( θ ) = display(frac(3, 5))$] #figure(figph[triangles], alt: "triangles", caption: none) (Answers may vary) #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 7 over 2")[$tan ( θ ) = display(frac(7, 2))$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 11 over 4")[$tan ( θ ) = display(frac(11, 4))$] #figure(figph[triangles], alt: "triangles", caption: none) (Answers may vary) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 9")[$sin ( θ ) = display(frac(4, 9))$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 9")[$sin ( θ ) = display(frac(1, 9))$] #figure(figph[triangles], alt: "triangles", caption: none) (Answers may vary) #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 7 over 8")[$cos ( θ ) = display(frac(7, 8))$] For Problems 17–22, use one of the three trigonometric ratios to find the unknown side of the triangle. Round your answer to hundredths. #figure(figph[triangle], alt: "triangle", caption: none) 14.41 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 37.86 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 86.08 #figure(figph[triangle], alt: "triangle", caption: none) For Problems 23–26, sketch and label a right triangle with the given properties. One angle is #math.equation(block: false, alt: "40 degrees")[$40^(∘)$], the side opposite that angle is 8 inches #figure(figph[triangle], alt: "triangle", caption: none) One angle is #math.equation(block: false, alt: "65 degrees")[$65^(∘)$], the side adjacent to that angle is 30 yards One angle is #math.equation(block: false, alt: "28 degrees")[$28^(∘)$], the hypotenuse is 56 feet #figure(figph[triangle], alt: "triangle", caption: none) One leg is 15 meters, the hypotenuse is 18 meters For Problems 27–34, + Sketch a right triangle that illustrates the situation. Label your sketch with the given information. + Choose the appropriate trig ratio and write an equation, then solve the problem. To measure the height of cloud cover, airport controllers fix a searchlight to shine a vertical beam on the clouds. The searchlight is 120 yards from the office. A technician in the office measures the angle of elevation to the light on the cloud cover at #math.equation(block: false, alt: "54.8 degrees")[$54.8^(∘)$]. What is the height of the cloud cover? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "tan ⁡ open parenthesis 54.8 degrees close parenthesis equals the fraction h over 20")[$tan ( 54.8^(∘) ) = display(frac(h, 20))$], 170.1 yd To measure the distance across a canyon, Evel first sights an interesting rock directly opposite on the other side. He then walks 200 yards down the rim of the canyon and sights the rock again, this time at an angle of #math.equation(block: false, alt: "18.5 degrees")[$18.5^(∘)$] from the canyon rim. What is the width of the canyon? A salvage ship is searching for the wreck of a pirate vessel on the ocean floor. Using sonar, they locate the wreck at an angle of depression of #math.equation(block: false, alt: "36.2 degrees")[$36.2^(∘)$]. The depth of the ocean at their location is 260 feet. How far should they move so that they are directly above the wrecked vessel? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "tan ⁡ open parenthesis 36.2 degrees close parenthesis equals the fraction 260 over d")[$tan ( 36.2^(∘) ) = display(frac(260, d))$], 355.2 ft Ramps for wheelchairs should be no steeper than an angle of #math.equation(block: false, alt: "6 degrees")[$6^(∘)$]. How much horizontal distance should be allowed for a ramp that rises 5 feet in height? The radio signal from a weather balloon indicates that it is 1500 meters from a meteorologist on the ground. The angle of elevation to the balloon is #math.equation(block: false, alt: "48 degrees")[$48^(∘)$]. What is the balloon's altitude? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis 48 degrees close parenthesis equals the fraction a over 1500")[$sin ( 48^(∘) ) = display(frac(a, 1500))$], 1114.7 m According to Chinese legend, around 200 BC the general Han Xin used a kite to determine the distance from his location to an enemy palace. He then dug a secret tunnel which emerged inside the palace. When the kite was directly above the palace, its angle of elevation was #math.equation(block: false, alt: "27 degrees")[$27^(∘)$] and the string to the kite was 1850 feet long. How far did Han Xin's troops have to dig? A cable car on a ski lift traverses a horizontal distance of 1800 meters at an angle of #math.equation(block: false, alt: "38 degrees")[$38^(∘)$]. How long is the cable? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "cos ⁡ open parenthesis 38 degrees close parenthesis equals the fraction 1800 over x")[$cos ( 38^(∘) ) = display(frac(1800, x))$], 2284.2 m Zelda is building the loft on her summer cottage. At its central point, the height of the loft is 8 feet, and the pitch of the roof should be #math.equation(block: false, alt: "24 degrees")[$24^(∘)$]. How long should the rafters be? For Problems 35–40, use a trig ratio to write an equation for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "θ")[$θ$]. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "x equals the fraction 82 over tan ⁡ open parenthesis θ close parenthesis")[$x = display(frac(82, tan ( θ )))$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "x equals 11 sin ⁡ open parenthesis θ close parenthesis")[$x = 11 " " sin ( θ )$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "x equals the fraction 9 over c o s open parenthesis θ close parenthesis")[$x = display(frac(9, c o s ( θ)) )$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 41–44, find the altitude of the triangle. Round your answer to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "36 sin ⁡ open parenthesis 25 degrees close parenthesis approximately equals 15.21")[$36 " " sin ( 25^(∘) ) ≈ 15.21$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "46 sin ⁡ open parenthesis 20 degrees close parenthesis approximately equals 15.73")[$46 " " sin ( 20^(∘) ) ≈ 15.73$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 45 and 46, find the length of the chord #math.equation(block: false, alt: "A B")[$A B$]. Round your answer to two decimal places. #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "12 sin ⁡ open parenthesis 40 degrees close parenthesis approximately equals 7.71")[$12 " " sin ( 40^(∘) ) ≈ 7.71$] #figure(figph[circle], alt: "circle", caption: none) For Problems 47–50, fill in the table. #figure(figph[triangle], alt: "triangle", caption: none) #figure(table( columns: 4, align: left, inset: 6pt, table.header([$" " " " " " " "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "θ")[$θ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], [#math.equation(block: false, alt: "ϕ")[$ϕ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([$" " " " " " " "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "the fraction 3 over 5")[$frac(3, 5)$]], [#math.equation(block: false, alt: "the fraction 4 over 5")[$frac(4, 5)$]], [#math.equation(block: false, alt: "the fraction 3 over 4")[$frac(3, 4)$]], [#math.equation(block: false, alt: "ϕ")[$ϕ$]], [#math.equation(block: false, alt: "the fraction 4 over 5")[$frac(4, 5)$]], [#math.equation(block: false, alt: "the fraction 3 over 5")[$frac(3, 5)$]], [#math.equation(block: false, alt: "the fraction 4 over 3")[$frac(4, 3)$]], )) #figure(figph[triangle], alt: "triangle", caption: none) #figure(table( columns: 4, align: left, inset: 6pt, table.header([$" " " " " " " "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "θ")[$θ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], [#math.equation(block: false, alt: "ϕ")[$ϕ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], )) #figure(figph[triangle], alt: "triangle", caption: none) #figure(table( columns: 4, align: left, inset: 6pt, table.header([$" " " " " " " "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "θ")[$θ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], [#math.equation(block: false, alt: "ϕ")[$ϕ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([$" " " " " " " "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 5")[$frac(1, sqrt(5))$]], [#math.equation(block: false, alt: "the fraction 2 over the square root of 5")[$frac(2, sqrt(5))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "ϕ")[$ϕ$]], [#math.equation(block: false, alt: "the fraction 2 over the square root of 5")[$frac(2, sqrt(5))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 5")[$frac(1, sqrt(5))$]], [#math.equation(block: false, alt: "2")[$2$]], )) #figure(figph[triangle], alt: "triangle", caption: none) #figure(table( columns: 4, align: left, inset: 6pt, table.header([$" " " " " " " "$], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]]), [#math.equation(block: false, alt: "θ")[$θ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], [#math.equation(block: false, alt: "ϕ")[$ϕ$]], [$" " " " " " " "$], [$" " " " " " " "$], [$" " " " " " " "$], )) + In each of the figures for Problems 47-50, what is the relationship between the angles #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "ϕ")[$ϕ$]? + Study the tables for Problems 47-50. What do you notice about the values of sine and cosine for the angles #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "ϕ")[$ϕ$]? Explain why this is true. + #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "ϕ")[$ϕ$] are complements. + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis ϕ close parenthesis")[$sin ( θ ) = cos ( ϕ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis ϕ close parenthesis")[$cos ( θ ) = sin ( ϕ )$]. The side opposite #math.equation(block: false, alt: "θ")[$θ$] is the side adjacent to #math.equation(block: false, alt: "ϕ")[$ϕ$], and vice versa. There is a relationship between the tangent, the sine, and the cosine of any angle. Study the tables for Problems 47-50 to discover this relationship. Write your answer as an equation. + Use the figure to explain what happens to #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] as #math.equation(block: false, alt: "θ")[$θ$] increases, and why. + Use the figure to explain what happens to #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] as #math.equation(block: false, alt: "θ")[$θ$] increases, and why. #figure(figph[Right triangles with a common base and increasing opposite leg], alt: "Right triangles with a common base and increasing opposite leg", caption: none) + As #math.equation(block: false, alt: "θ")[$θ$] increases, #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] increases also. The side opposite #math.equation(block: false, alt: "θ")[$θ$] increases in length while the side adjacent to #math.equation(block: false, alt: "θ")[$θ$] remains fixed. + As #math.equation(block: false, alt: "θ")[$θ$] increases, #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] decreases. The side adjacent to #math.equation(block: false, alt: "θ")[$θ$] remains fixed while the hypotenuse increases in length. + Fill in the table for values of #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Round your answers to four decimal places.#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: "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")[$" " 80^(∘)$]]), [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], )) + Fill in the table for values of #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Round your answers to three decimal places.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "81 degrees")[$" " 81^(∘)$]], [#math.equation(block: false, alt: "82 degrees")[$" " 82^(∘)$]], [#math.equation(block: false, alt: "83 degrees")[$" " 83^(∘)$]], [#math.equation(block: false, alt: "84 degrees")[$" " 84^(∘)$]], [#math.equation(block: false, alt: "85 degrees")[$" " 85^(∘)$]], [#math.equation(block: false, alt: "86 degrees")[$" " 86^(∘)$]], [#math.equation(block: false, alt: "87 degrees")[$" " 87^(∘)$]], [#math.equation(block: false, alt: "88 degrees")[$" " 88^(∘)$]], [#math.equation(block: false, alt: "89 degrees")[$" " 89^(∘)$]]), [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], )) + What happens to #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] as #math.equation(block: false, alt: "θ")[$θ$] increases? + What value does your calculator give for #math.equation(block: false, alt: "tan ⁡ open parenthesis 90 degrees close parenthesis")[$tan ( 90^(∘) )$]? Why? Explain why it makes sense that #math.equation(block: false, alt: "sin ⁡ open parenthesis 0 degrees close parenthesis equals 0")[$sin ( 0^(∘) ) = 0$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 90 degrees close parenthesis equals 1")[$sin ( 90^(∘) ) = 1$]. Use a figure to illustrate your explanation. As #math.equation(block: false, alt: "θ")[$θ$] decreases toward #math.equation(block: false, alt: "0 degrees")[$0^(∘)$], the side opposite #math.equation(block: false, alt: "θ")[$θ$] approaches a length of 0, so sin #math.equation(block: false, alt: "open parenthesis θ close parenthesis")[$( θ )$] approaches 0. But as #math.equation(block: false, alt: "θ")[$θ$] increases toward #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], the length of the side opposite #math.equation(block: false, alt: "θ")[$θ$] approaches the length of the hypotenuse, so #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] approaches 1. Explain why it makes sense that #math.equation(block: false, alt: "cos ⁡ open parenthesis 0 degrees close parenthesis equals 1")[$cos ( 0^(∘) ) = 1$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 90 degrees close parenthesis equals 0")[$cos ( 90^(∘) ) = 0$]. Use a figure to illustrate your explanation For Problems 57–60, explain why the trigonometric ratio is #emph[not] correct. #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 9")[$sin ( θ ) = display(frac(5, 9))$] #figure(figph[triangle], alt: "triangle", caption: none) The triangle is not a right tringle. #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 7")[$tan ( θ ) = display(frac(4, 7))$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 21 over 20")[$cos ( θ ) = display(frac(21, 20))$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: false, alt: "the fraction 21 over 20")[$display(frac(21, 20))$] is the ratio of hypotenuse to the adjacent side, which is the reciprocal of #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 8 over 10")[$sin ( θ ) = display(frac(8, 10))$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 61–64, sketch and label a right triangle, then fill in the blank. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.2358")[$sin ( θ ) = 0.2358$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis 90 degrees minus θ close parenthesis equals bar")[$cos ( 90^(∘) − θ ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction 3 over 11")[$cos ( α ) = display(frac(3, 11))$], then #math.equation(block: false, alt: "bar open parenthesis 90 degrees minus α close parenthesis equals the fraction 3 over 11")[$underline(#h(3em)) ( 90^(∘) − α ) = display(frac(3, 11))$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis 42 degrees close parenthesis equals n")[$sin ( 42^(∘) ) = n$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals n")[$cos ( underline(#h(3em)) ) = n$]. + If #math.equation(block: false, alt: "cos ⁡ open parenthesis 13 degrees close parenthesis equals z")[$cos ( 13^(∘) ) = z$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis equals z")[$sin ( underline(#h(3em)) ) = z$]. + 0.2358 + sine + #math.equation(block: false, alt: "48 degrees")[$48^(∘)$] + #math.equation(block: false, alt: "77 degrees")[$77^(∘)$] + If #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis equals the fraction 2 over the square root of 7")[$cos ( β ) = display(frac(2, sqrt(7)))$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis 90 degrees minus β close parenthesis equals bar")[$sin ( 90^(∘) − β ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals 0.693")[$sin ( ϕ ) = 0.693$], then \_\_\_\_\_\_#math.equation(block: false, alt: "open parenthesis 90 degrees minus ϕ close parenthesis equals 0.693")[$( 90^(∘) − ϕ ) = 0.693$]. + If #math.equation(block: false, alt: "cos ⁡ open parenthesis 87 degrees close parenthesis equals p")[$cos ( 87^(∘) ) = p$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis bar close parenthesis equals p")[$sin ( underline(#h(3em)) ) = p$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis 59 degrees close parenthesis equals w")[$sin ( 59^(∘) ) = w$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis bar close parenthesis equals w")[$cos ( underline(#h(3em)) ) = w$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the fraction 5 over 13")[$sin ( ϕ ) = display(frac(5, 13))$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis ϕ close parenthesis equals the fraction 12 over 13")[$cos ( ϕ ) = display(frac(12, 13))$], then #math.equation(block: false, alt: "tan ⁡ open parenthesis ϕ close parenthesis equals bar")[$tan ( ϕ ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis equals the fraction 1 over the square root of 10")[$cos ( β ) = display(frac(1, sqrt(10)))$], and #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis equals the fraction 3 over the square root of 10")[$sin ( β ) = display(frac(3, sqrt(10)))$], then #math.equation(block: false, alt: "tan ⁡ open parenthesis β close parenthesis equals bar")[$tan ( β ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "tan ⁡ open parenthesis B close parenthesis equals the fraction 2 over the square root of 5")[$tan ( B ) = display(frac(2, sqrt(5)))$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis B close parenthesis equals the fraction the square root of 5 over 3")[$cos ( B ) = display(frac(sqrt(5), 3))$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis B close parenthesis equals bar")[$sin ( B ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis W close parenthesis equals the square root of the fraction 3 over 7")[$sin ( W ) = sqrt(display(frac(3, 7)))$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis W close parenthesis equals the fraction the square root of 3 over 2")[$tan ( W ) = display(frac(sqrt(3), 2))$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis W close parenthesis equals bar")[$cos ( W ) = underline(#h(3em))$]. + #math.equation(block: false, alt: "the fraction 5 over 12")[$display(frac(5, 12))$] + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] + #math.equation(block: false, alt: "the fraction 2 over the square root of 7")[$display(frac(2, sqrt(7)))$] + If #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 2 over the square root of 10")[$cos ( θ ) = display(frac(2, sqrt(10)))$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the square root of the fraction 3 over 5")[$sin ( θ ) = sqrt(display(frac(3, 5)))$], then #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals bar")[$tan ( θ ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction the square root of 2 over 4")[$sin ( α ) = display(frac(sqrt(2), 4))$], and #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction the square root of 14 over 4")[$cos ( α ) = display(frac(sqrt(14), 4))$], then #math.equation(block: false, alt: "tan ⁡ open parenthesis α close parenthesis equals bar")[$tan ( α ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "tan ⁡ open parenthesis A close parenthesis equals the fraction the square root of 7 over 3")[$tan ( A ) = display(frac(sqrt(7), 3))$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis A close parenthesis equals the fraction 3 over 4")[$cos ( A ) = display(frac(3, 4))$], then #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals bar")[$sin ( A ) = underline(#h(3em))$]. + If #math.equation(block: false, alt: "sin ⁡ open parenthesis V close parenthesis equals the fraction the square root of 10 over 5")[$sin ( V ) = display(frac(sqrt(10), 5))$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis V close parenthesis equals the square root of the fraction 2 over 3")[$tan ( V ) = sqrt(display(frac(2, 3)))$], then #math.equation(block: false, alt: "cos ⁡ open parenthesis V close parenthesis equals bar")[$cos ( V ) = underline(#h(3em))$]. Explain why the cosine of a #math.equation(block: false, alt: "73 degrees")[$73^(∘)$] angle is always the same, no matter what size triangle the angle is in. Illustrate your explanation with a sketch. Although the triangles may differ in size, the ratio of the side adjacent to the angle to the hypotenuse of the triangle remains the same because the triangles would all be similar, and hence corresponding sides are proportional. + Use your calculator to fill in a table of values for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$], rounded to hundredths.#figure(table( columns: 8, 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: "15 degrees")[$" " 15^(∘)$]], [#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: "75 degrees")[$" " 75^(∘)$]], [#math.equation(block: false, alt: "90 degrees")[$" " 90^(∘)$]]), [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], )) + If you plotted the points in your table, would they lie on a straight line? Why or why not? + What is the slope of the line through the origin and point #math.equation(block: false, alt: "P")[$P$]? + What is the tangent of the angle #math.equation(block: false, alt: "θ")[$θ$]? + On the same grid, sketch an angle whose tangent is #math.equation(block: false, alt: "the fraction 8 over 5.")[$display(frac(8, 5)) .$] #figure(figph[grid], alt: "grid", caption: none) + #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] + #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] + #figure(figph[triangle], alt: "triangle", caption: none) + Use your calculator to complete the table. Rounded your answers to hundredths.#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "14 degrees")[$" " 14^(∘)$]], [#math.equation(block: false, alt: "22 degrees")[$" " 22^(∘)$]], [#math.equation(block: false, alt: "35 degrees")[$" " 35^(∘)$]], [#math.equation(block: false, alt: "42 degrees")[$" " 42^(∘)$]], [#math.equation(block: false, alt: "58 degrees")[$" " 58^(∘)$]], [#math.equation(block: false, alt: "78 degrees")[$" " 78^(∘)$]]), [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], [$" " " " " "$], )) + Use the values of #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] to sketch all the angles listed in the table. Locate the vertex of each angle at the origin, and the initial side along the positive #math.equation(block: false, alt: "x")[$x$]-axis. #figure(figph[triangle], alt: "triangle", caption: none)