#set document(title: "2.3 Solving Right Triangles", 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")) == 2.3#h(0.6em)Solving Right Triangles === Introduction A triangle has six parts: three sides and three angles. In a right triangle, we know that one of the angles is #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. If we know three parts of a right triangle, including one of the sides, we can use trigonometry to find all the other unknown parts. This is called #emph[solving] the triangle. #examplebox("Example 1")[][ A 10 foot ladder is leaning againt a wall. If the base of the ladder is too far from the wall, the base can slide out. If the ladder base is too close to the wall, there’s a risk that the ladder could tip over backwards. For safety reasons, the leaning ladder should make a #math.equation(block: false, alt: "75 degrees")[$75^(∘)$] angle with the ground. + How far should the base of the ladder be from the wall? + How far up the wall will the top of the ladder reach? #figure(figph[ladder], alt: "ladder", caption: none) #solutionbox[ + The distance between the base of the ladder and the wall is the side adjacent to the #math.equation(block: false, alt: "75 degrees")[$75^(∘)$] angle. We can find the side adjacent to the #math.equation(block: false, alt: "75 degrees")[$75^(∘)$] angle angle by using the cosine ratio. #math.equation(block: true, alt: "cos ⁡ open parenthesis 75 degrees close parenthesis, equals the fraction adjacent over hypotenuse; 0.2588, equals the fraction b over 10; b, equals 10 open parenthesis 0.2588 close parenthesis equals 2.588")[$cos ( 75^(∘) ) & = display(frac("adjacent", "hypotenuse")) \ 0.2588 & = display(frac(b, 10)) \ b & = 10 ( 0.2588 ) = 2.588$] The base of the ladder should be about 2.6 feet from the wall. + To find side how far up the wall the ladder will reach, we could use the Pythagorean theorem now, but it is better to use given information, rather than values we have calculated, to find the other unknown parts. We will use the sine ratio. #math.equation(block: true, alt: "sin ⁡ open parenthesis 75 degrees close parenthesis, equals the fraction opposite over hypotenuse; 0.9659, equals the fraction a over 10; b, equals 10 open parenthesis 0.9659 close parenthesis equals 9.659")[$sin ( 75^(∘) ) & = display(frac("opposite", "hypotenuse")) \ 0.9659 & = display(frac(a, 10)) \ b & = 10 ( 0.9659 ) = 9.659$] The ladder will reach about 9.7 feet. ] ] Sketch a right triangle with - one angle of #math.equation(block: false, alt: "37 degrees")[$37^(∘)$], - the side adjacent to that angle of length 5 centimeters. Without doing the calculations, list the steps you would use to solve the triangle. Use #math.equation(block: false, alt: "tan ⁡ open parenthesis 37 degrees close parenthesis")[$tan ( 37^(∘) )$] to find the opposite side. Use #math.equation(block: false, alt: "cos ⁡ open parenthesis 37 degrees close parenthesis")[$cos ( 37^(∘) )$] to find the hypotenuse. Subtract #math.equation(block: false, alt: "37 degrees")[$37^(∘)$] from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] to find the third angle. === Finding an Angle While watching her niece at the playground, Francine wonders how steep the slide is. She happens to have a tape measure and her calculator with her, and finds that the slide is 77 inches high and covers a horizontal distance of 136 inches, as shown below. #figure(figph[slide with triangle], alt: "slide with triangle", caption: none) Francine knows that one way to describe the steepness of an incline is to calculate its slope, which in this case is #math.equation(block: true, alt: "the fraction Δ y over Δ x equals the fraction 77 over 136 equals 0.5662")[$display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(77, 136)) = 0.5662$] However, Francine would really like to know what angle the slide makes with the horizontal. She realizes that the slope she has just calculated is also the tangent of the angle she wants. If we know the tangent of an angle, can we find the angle? Yes, we can: locate the key labeled #math.equation(block: false, alt: "tan to the power minus 1")[$tan^(− 1)$] on your calculator; it is probably the second function above the TAN key. Enter $#h(2em) #h(2em) #h(2em)$2nd TAN 0.5662 and you should find that #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis 0.5662 close parenthesis equals 29.52 degrees")[$tan^(− 1) ( 0.5662 ) = 29.52^(∘)$] . This means that #math.equation(block: false, alt: "29.52 degrees")[$29.52^(∘)$] is the angle whose tangent is #math.equation(block: false, alt: "0.5662")[$0.5662$]. We read the notation as "#strong[inverse tangent] of 0.5662 is 29.52 degrees." When we find #math.equation(block: false, alt: "tan to the power minus 1")[$" " tan^(− 1)$] of a number, we are finding an angle whose tangent is that number. Similarly, #math.equation(block: false, alt: "sin to the power minus 1")[$" " sin^(− 1)$] and #math.equation(block: false, alt: "cos to the power minus 1")[$" " cos^(− 1)$] are read as “inverse sine” and “inverse cosine.” They find an angle with the given sine or cosine. #examplebox("Example 2")[][ Find the angle whose sine is #math.equation(block: false, alt: "0.6834")[$0.6834$]. #solutionbox[ Enter 2nd SIN 0.6834 into your calculator to find #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis 0.6834 close parenthesis equals 43.11 degrees")[$sin^(− 1) ( 0.6834 ) = 43.11^(∘)$] So #math.equation(block: false, alt: "43.11 degrees")[$43.11^(∘)$] is the angle whose sine is #math.equation(block: false, alt: "0.6834")[$0.6834$]. Or we can say that #math.equation(block: true, alt: "sin ⁡ open parenthesis 43.11 degrees close parenthesis equals 0.6834")[$sin ( 43.11^(∘) ) = 0.6834$] You can check the last equation on your calculator. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the last example, the two equations #math.equation(block: true, alt: "sin ⁡ open parenthesis 43.11 degrees close parenthesis equals 0.6834 and sin to the power minus 1 ⁡ open parenthesis 0.6834 close parenthesis equals 43.11 degrees")[$sin ( 43.11^(∘) ) = 0.6834 " " " " " " " " " " " " "and" " " " " " " " " " " " " sin^(− 1) ( 0.6834 ) = 43.11^(∘)$] say the same thing in different ways. ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The notation #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$] does #emph[not] mean #math.equation(block: false, alt: "the fraction 1 over sin ⁡ open parenthesis x close parenthesis")[$display(frac(1, sin ( x )))$]. It is true that we use negative exponents to indicate reciprocals of numbers, for example #math.equation(block: false, alt: "a to the power minus 1 equals the fraction 1 over a")[$a^(− 1) = display(frac(1, a))$] and #math.equation(block: false, alt: "3 to the power minus 1 equals the fraction 1 over 3")[$3^(− 1) = display(frac(1, 3))$]. But “sin” by itself is not a variable. - #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$] means “the angle whose sine is #math.equation(block: false, alt: "x")[$x$]” - #math.equation(block: false, alt: "the fraction 1 over sin ⁡ open parenthesis x close parenthesis")[$display(frac(1, sin ( x )))$] means “the reciprocal of the sine of angle #math.equation(block: false, alt: "x")[$x$]” (You may recall that #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$f^(− 1) ( x )$] denotes the inverse function for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]. We will study trigonometric functions in Chapter 4.) ] Write the following fact in two different ways: #math.equation(block: false, alt: "68 degrees")[$68^(∘)$] is the angle whose cosine is #math.equation(block: false, alt: "0.3746")[$0.3746$]. #math.equation(block: false, alt: "cos ⁡ open parenthesis 68 degrees close parenthesis equals 0.3746")[$" " cos ( 68^(∘) ) = 0.3746 " "$] or #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis 0.3746 close parenthesis equals 68 degrees")[$" " cos^(− 1) ( 0.3746 ) = 68^(∘)$] #examplebox("Example 3")[][ Find the angle of inclination of a hill if you gain 400 feet in elevation while traveling half a mile. #solutionbox[ A sketch of the hill is shown at right. (Recall that 1 mile = 5280 feet.) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis, equals the fraction 400 over 2640 equals 0.15 bar; θ, equals sin to the power minus 1 ⁡ open parenthesis 0.15 bar close parenthesis equals 8.71 degrees")[$sin ( θ ) & = display(frac(400, 2640)) = 0. limits(15)^(―) \ θ & = sin^(− 1) ( 0. limits(15)^(―) ) = 8.71^(∘)$] #figure(figph[right triangle for hill], alt: "right triangle for hill", caption: none) The angle of inclination of the hill is about #math.equation(block: false, alt: "8.7 degrees")[$8.7^(∘)$]. ] ] The tallest living tree is a coast redwood named Hyperion, at 378.1 feet tall. If you stand 100 feet from the base of the tree, what is the angle of elevation of your line of sight to the top of the tree? Round your answer to the nearest degree. #figure(figph[Figure], alt: "Figure", caption: none) #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis the fraction 378.1 over 100 close parenthesis equals 75 degrees")[$tan^(− 1) ( display(frac(378.1, 100)) ) = 75^(∘)$] === The Special Angles The trigonometric ratios for most angles are irrational numbers, but there are a few angles whose trig ratios are “nice” values. You already know one of these values: the sine of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]. Because the sides of a right triangle are related by the Pythagorean theorem, if we know any one of the trig ratios for an angle, we can find the others. Recall that the side opposite a #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angle is half the length of the hypotenuse, so #math.equation(block: false, alt: "sin ⁡ open parenthesis 30 degrees close parenthesis equals the fraction 1 over 2")[$sin ( 30^(∘) ) = display(frac(1, 2))$]. The figure at right shows a 30-60-90 triangle with hypotenuse of length 2. The opposite side has length 1, and we can calculate the length of the adjacent side. #math.equation(block: true, alt: "1 squared plus b squared, equals 2 squared; b squared, equals 2 squared minus 1 squared equals 3; b, equals the square root of 3")[$1^(2) + b^(2) & = 2^(2) \ b^(2) & = 2^(2) − 1^(2) = 3 \ b & = sqrt(3)$] #figure(figph[triangle], alt: "triangle", caption: none) Now we know the cosine and tangent of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis 30 degrees close parenthesis equals the fraction adjacent over hypotenuse equals the fraction the square root of 3 over 2 tan ⁡ open parenthesis 30 degrees close parenthesis equals the fraction opposite over adjacent equals the fraction 1 over the square root of 3")[$cos ( 30^(∘) ) = display(frac("adjacent", "hypotenuse")) = display(frac(sqrt(3), 2)) #h(2em) tan ( 30^(∘) ) = display(frac("opposite", "adjacent")) = display(frac(1, sqrt(3)))$] These are #emph[exact] values for the trig ratios, but we can also find decimal approximations. Use your calculator to verify the following approximate values. #math.equation(block: true, alt: "exact value, approximation; cos ⁡ open parenthesis 30 degrees close parenthesis, equals the fraction the square root of 3 over 2, approximately equals 0.8660; tan ⁡ open parenthesis 30 degrees close parenthesis, equals the fraction 1 over the square root of 3, approximately equals 0.5774")[$& "exact value" & & "approximation" \ cos ( 30^(∘) ) & = display(frac(sqrt(3), 2)) & & ≈ 0.8660 \ tan ( 30^(∘) ) & = display(frac(1, sqrt(3))) & & ≈ 0.5774$] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ It is important for you to understand the difference between exact and approximate values. These decimal approximations, like nearly all the other trig values your calculator gives you, are rounded off. Even if your calculator shows you ten or twelve digits, the values are not exactly correct -- although they are quite adequate for most practical calculations! ] The angles #math.equation(block: false, alt: "30 degrees")[$30^(∘)$], #math.equation(block: false, alt: "60 degrees")[$60^(∘)$], and #math.equation(block: false, alt: "45 degrees")[$45^(∘)$] are “special” because we can easily find exact values for their trig ratios, and use those exact values to find exact lengths for the sides of triangles with those angles. #examplebox("Example 4")[][ Matterhorn Chocolate bars are sold in boxes shaped like triangular prisms. The two triangular ends of a box are equilateral triangles with 8 centimeter sides. When stacking the boxes on a shelf, the second level will be a triangle's altitude above bottom level. Find the exact length of the triangle's altitude, #math.equation(block: false, alt: "h")[$h$]. #solutionbox[ The altitude divides the triangle into two 30-60-90 right triangles as shown in the figure. The altitude is adjacent to the #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angle, and the hypotenuse of the right triangle is 8 centimeters. Thus, #figure(figph[equilateral triangle], alt: "equilateral triangle", caption: none) #math.equation(block: true, alt: "cos ⁡ open parenthesis 30 degrees close parenthesis, equals the fraction adjacent over hypotenuse, Fill in the values.; the fraction the square root of 3 over 2, equals the fraction h over 8, Multiply both sides by 8.; h, equals 8 open parenthesis the fraction the square root of 3 over 2 close parenthesis equals 4 the square root of 3")[$cos ( 30^(∘) ) & = display(frac("adjacent", "hypotenuse")) & & "Fill in the values." \ display(frac(sqrt(3), 2)) & = display(frac(h, 8)) & & "Multiply both sides by 8." \ h & = 8 ( display(frac(sqrt(3), 2)) ) = 4 sqrt(3)$] The altitude is #math.equation(block: false, alt: "4 the square root of 3")[$4 sqrt(3)$] centimeters long. From this exact answer, we can find approximations to any degree of accuracy we like. You can check that #math.equation(block: false, alt: "4 the square root of 3 approximately equals 6.9282")[$4 sqrt(3) ≈ 6.9282$], so the altitude is approximately 6.9 centimeters long. ] ] Use the figure in the previous example to find exact values for the sine, cosine, and tangent of #math.equation(block: false, alt: "60 degrees")[$60^(∘)$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis equals the fraction the square root of 3 over 2")[$sin ( 60^(∘) ) = display(frac(sqrt(3), 2))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 60 degrees close parenthesis equals the fraction 1 over 2")[$" " " " cos ( 60^(∘) ) = display(frac(1, 2))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis 60 degrees close parenthesis equals the square root of 3")[$" " " " tan ( 60^(∘) ) = sqrt(3)$] There is one more special angle: #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]. We find the trig ratios for this angle using an isosceles right triangle. Because the base angles of an isosceles triangle are equal, they must both be #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]. The figure shows an isosceles right triangle with equal sides of length 1. You can use the Pythagorean theorem to show that the hypotenuse has length #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$], so the trig ratios for #math.equation(block: false, alt: "45 degrees")[$45^(∘)$] are #math.equation(block: true, alt: "sin ⁡ open parenthesis 45 degrees close parenthesis, equals the fraction opposite over hypotenuse equals the fraction 1 over the square root of 2 approximately equals 0.7071; cos ⁡ open parenthesis 45 degrees close parenthesis, equals the fraction adjacent over hypotenuse equals the fraction 1 over the square root of 2 approximately equals 0.7071; tan ⁡ open parenthesis 45 degrees close parenthesis, equals the fraction opposite over adjacent equals 1")[$sin ( 45^(∘) ) & = display(frac("opposite", "hypotenuse")) = display(frac(1, sqrt(2))) ≈ 0.7071 \ cos ( 45^(∘) ) & = display(frac("adjacent", "hypotenuse")) = display(frac(1, sqrt(2))) ≈ 0.7071 \ tan ( 45^(∘) ) & = display(frac("opposite", "adjacent")) = 1$] #figure(figph[45-45-90 triangle], alt: "45-45-90 triangle", caption: none) === The Trigonometric Ratios for the Special Angles Here is a summary of the trig ratios for the special angles. #figure(table( columns: 4, align: left, inset: 6pt, [Trigonometric Ratios for the Special Angles], [], [], [], [Angle], [Sine], [Cosine], [Tangent], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "the fraction 1 over 2 equals 0.5")[$display(frac(1, 2)) = 0.5$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2 approximately equals 0.8660")[$display(frac(sqrt(3), 2)) ≈ 0.8660$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3 approximately equals 0.5774")[$display(frac(1, sqrt(3))) ≈ 0.5774$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2 approximately equals 0.7071")[$display(frac(1, sqrt(2))) ≈ 0.7071$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2 approximately equals 0.7071")[$display(frac(1, sqrt(2))) ≈ 0.7071$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2 approximately equals 0.8660")[$display(frac(sqrt(3), 2)) ≈ 0.8660$]], [#math.equation(block: false, alt: "the fraction 1 over 2 equals 0.5")[$display(frac(1, 2)) = 0.5$]], [#math.equation(block: false, alt: "the square root of 3 approximately equals 1.732")[$sqrt(3) ≈ 1.732$]], )) You should be familiar with the exact values for these trig ratios. A good way to remember them is to know the two special triangles shown below. From these triangles, you can always write down the three trig ratios for the special angles. #figure(figph[45-45-90 triangle], alt: "45-45-90 triangle", caption: none) #figure(figph[30-60-90 triangle], alt: "30-60-90 triangle", caption: none) You should also be able to recognize their decimal approximations. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We can use the special angles as benchmarks for estimating and mental calculation. For example, we know that #math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis approximately equals 0.8660")[$" " sin ( 60^(∘) ) ≈ 0.8660$], so if #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.95")[$" " sin ( θ ) = 0.95$] for some unknown angle #math.equation(block: false, alt: "θ")[$θ$], we know that #math.equation(block: false, alt: "θ greater than 60 degrees")[$" " θ > 60^(∘)$], because as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] the sine of #math.equation(block: false, alt: "θ")[$θ$] increases also. ] #examplebox("Example 5")[][ If #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis greater than the fraction the square root of 3 over 2")[$" " cos ( α ) > display(frac(sqrt(3), 2))$], what can we say about #math.equation(block: false, alt: "α")[$α$]? #solutionbox[ As an angle increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] to #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], its cosine decreases. Now, #math.equation(block: false, alt: "cos ⁡ open parenthesis 30 degrees close parenthesis equals the fraction the square root of 3 over 2")[$" " cos ( 30^(∘) ) = display(frac(sqrt(3), 2))$], so if #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis greater than the fraction the square root of 3 over 2")[$" " cos ( α ) > display(frac(sqrt(3), 2))$], then #math.equation(block: false, alt: "α")[$α$] must be less than #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]. ] ] If #math.equation(block: false, alt: "1 less than tan ⁡ open parenthesis β close parenthesis less than the square root of 3")[$" " 1 < tan ( β ) < sqrt(3)$], what can we say about #math.equation(block: false, alt: "β")[$β$]? Because #math.equation(block: false, alt: "tan ⁡ open parenthesis β close parenthesis")[$tan ( β )$] increases as #math.equation(block: false, alt: "β")[$β$] increases, we know that #math.equation(block: false, alt: "45 degrees less than β less than 60 degrees")[$" " 45^(∘) < β < 60^(∘)$]. Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify. 1. #math.equation(block: false, alt: "the square root of 2 the square root of 2")[$#h(0.222em) sqrt(2) sqrt(2)$] 2. #math.equation(block: false, alt: "the fraction 3 over the square root of 3")[$#h(0.222em) display(frac(3, sqrt(3)))$] 3. #math.equation(block: false, alt: "the square root of 8")[$#h(0.222em) sqrt(8)$] 4. #math.equation(block: false, alt: "the square root of the fraction 3 over 4")[$#h(0.222em) sqrt(display(frac(3, 4)))$] Rationalize the denominator. 5. #math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$#h(0.222em) display(frac(1, sqrt(2)))$] 6. #math.equation(block: false, alt: "the fraction 2 over the square root of 3")[$#h(0.222em) display(frac(2, sqrt(3)))$] 7. #math.equation(block: false, alt: "the fraction 6 over the square root of 3")[$#h(0.222em) display(frac(6, sqrt(3)))$] 8. #math.equation(block: false, alt: "the fraction 4 over the square root of 8")[$#h(0.222em) display(frac(4, sqrt(8)))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "2")[$2 #hide($display(frac(sqrt(3), 2))$)$] + #math.equation(block: false, alt: "the square root of 3")[$sqrt(3) #hide($#hide($display(frac(sqrt(3), 2))$)$)$] + #math.equation(block: false, alt: "2 the square root of 2")[$2 sqrt(2) #hide($display(frac(sqrt(3), 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: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$] + #math.equation(block: false, alt: "the fraction 2 the square root of 3 over 3")[$display(frac(2 sqrt(3), 3))$] + #math.equation(block: false, alt: "2 the square root of 3")[$2 sqrt(3) #hide($display(frac(sqrt(3), 2))$)$] + #math.equation(block: false, alt: "the square root of 2")[$sqrt(2) #hide($display(frac(sqrt(3), 2))$)$] ] === Section 2.3 Summary ==== Vocabulary - Solve a triangle - Inverse sine - Inverse cosine - Inverse tangent - Special angles - Exact value - Decimal approximation ==== Concepts + If we know one of the sides of a right triangle and any one of the other four parts, we can use trigonometry to find all the other unknown parts. + If we know one of the trigonometric ratios of an acute angle, we can find the angle using the inverse trig key on a calculator. + The exact values of trigonometric ratios of the special angles should be memorized.#figure(table( columns: 4, align: left, inset: 6pt, table.header([Angle], [Sine], [Cosine], [Tangent]), [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "the fraction 1 over 2 equals 0.5")[$display(frac(1, 2)) = 0.5$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2 approximately equals 0.8660")[$display(frac(sqrt(3), 2)) ≈ 0.8660$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3 approximately equals 0.5774")[$display(frac(1, sqrt(3))) ≈ 0.5774$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2 approximately equals 0.7071")[$display(frac(1, sqrt(2))) ≈ 0.7071$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2 approximately equals 0.7071")[$display(frac(1, sqrt(2))) ≈ 0.7071$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2 approximately equals 0.8660")[$display(frac(sqrt(3), 2)) ≈ 0.8660$]], [#math.equation(block: false, alt: "the fraction 1 over 2 equals 0.5")[$display(frac(1, 2)) = 0.5$]], [#math.equation(block: false, alt: "the square root of 3 approximately equals 1.732")[$sqrt(3) ≈ 1.732$]], )) + You can remember the trig values for the special angles if you know two triangles:#figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) + For the trigonometric ratios of most angles, your calculator gives approximations, not exact values. ==== Study Questions + How many parts of a right triangle (including the right angle) do you need to know in order to solve the triangle? + Why is it better to use the given values when solving a triangle, rather than values you have calculated? + What is the #math.equation(block: false, alt: "sin to the power minus 1")[$sin^(− 1)$] (or #math.equation(block: false, alt: "cos to the power minus 1")[$cos^(− 1)$] or #math.equation(block: false, alt: "tan to the power minus 1")[$tan^(− 1)$]) button on the calculator used for? + Which are the "special" angles, and why are they special? ==== Skills Practice each skill in the Homework Problems listed. + Solve a right triangle \#1-16, 63-74 + Use inverse trig ratio notation \#17-34 + Use trig ratios to find an angle \#17-22, 35-38 + Solve problems involving right triangles \#35-48 + Know the trig ratios for the special angles \#49-62, 75-78 === Homework 2.3 In these Homework Problems, we use the following standard notation for a right triangle: in #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], #math.equation(block: false, alt: "∠ C")[$upright(∠) C$] is a right angle. The side opposite #math.equation(block: false, alt: "∠ C")[$upright(∠) C$] has length #math.equation(block: false, alt: "c")[$c$], and so on. (See the figure at right.) #figure(figph[triangle], alt: "triangle", caption: none) For Problems 1–4, solve the triangle. Round answers to hundredths. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "A equals 61 degrees , a equals 25.26 , c equals 28.88")[$A = 61^(∘) , " " a = 25.26 , " " c = 28.88$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "A equals 68 degrees , a equals 0.93 , b equals 0.37")[$A = 68^(∘) , " " a = 0.93 , " " b = 0.37$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 5–10, + Sketch the right triangle described. + Solve the triangle. #math.equation(block: true, alt: "A equals 42 degrees , c equals 26")[$A = 42^(∘) , " " c = 26$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "B equals 48 degrees")[$B = 48^(∘)$], #math.equation(block: false, alt: "a equals 17.4")[$" " a = 17.4$], #math.equation(block: false, alt: "b equals 19.3")[$" " b = 19.3$] #math.equation(block: true, alt: "B equals 28 degrees , c equals 6.8")[$B = 28^(∘) , " " c = 6.8$] #math.equation(block: true, alt: "B equals 33 degrees , a equals 300")[$B = 33^(∘) , " " a = 300$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "A equals 57 degrees")[$A = 57^(∘)$], #math.equation(block: false, alt: "b equals 194.4")[$" " b = 194.4$], #math.equation(block: false, alt: "c equals 357.7")[$" " c = 357.7$] #math.equation(block: true, alt: "B equals 79 degrees , a equals 116")[$B = 79^(∘) , " " a = 116$] #math.equation(block: true, alt: "A equals 12 degrees , a equals 4")[$A = 12^(∘) , " " a = 4$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "B equals 78 degrees")[$B = 78^(∘)$], #math.equation(block: false, alt: "b equals 18.8")[$" " b = 18.8$], #math.equation(block: false, alt: "c equals 19.2")[$" " c = 19.2$] #math.equation(block: true, alt: "A equals 50 degrees , a equals 10")[$A = 50^(∘) , " " a = 10$] For Problems 11–16, + Sketch the right triangle described. + Without doing the calculations, list the steps you would use to solve the triangle. #math.equation(block: true, alt: "B equals 53.7 degrees , b equals 8.2")[$B = 53.7^(∘) , " " b = 8.2$] + #figure(figph[triangle], alt: "triangle", caption: none) + - Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis 53.7 degrees close parenthesis equals the fraction 8.2 over c")[$sin ( 53.7^(∘) ) = display(frac(8.2, c))$] for #math.equation(block: false, alt: "c")[$c$]. - Solve #math.equation(block: false, alt: "tan ⁡ open parenthesis 53.7 degrees close parenthesis equals the fraction 8.2 over a")[$tan ( 53.7^(∘) ) = display(frac(8.2, a))$] for #math.equation(block: false, alt: "a")[$a$]. - Subtract #math.equation(block: false, alt: "53.7 degrees")[$53.7^(∘)$] from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] to find #math.equation(block: false, alt: "A")[$A$]. #math.equation(block: true, alt: "B equals 80 degrees , a equals 250")[$B = 80^(∘) , " " a = 250$] #math.equation(block: true, alt: "A equals 25 degrees , b equals 40")[$A = 25^(∘) , " " b = 40$] + #figure(figph[triangle], alt: "triangle", caption: none) + - Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis 25 degrees close parenthesis equals the fraction 40 over c")[$cos ( 25^(∘) ) = display(frac(40, c))$] for #math.equation(block: false, alt: "c")[$c$]. - Solve #math.equation(block: false, alt: "tan ⁡ open parenthesis 25 degrees close parenthesis equals the fraction a over 40")[$tan ( 25^(∘) ) = display(frac(a, 40))$] for #math.equation(block: false, alt: "a")[$a$]. - Subtract #math.equation(block: false, alt: "25 degrees")[$25^(∘)$] from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] to find #math.equation(block: false, alt: "B")[$B$]. #math.equation(block: true, alt: "A equals 15 degrees , c equals 62")[$A = 15^(∘) , " " c = 62$] #math.equation(block: true, alt: "A equals 64.5 degrees , c equals 24")[$A = 64.5^(∘) , " " c = 24$] + #figure(figph[triangle], alt: "triangle", caption: none) + - Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis 64.5 degrees close parenthesis equals the fraction a over 24")[$sin ( 64.5^(∘) ) = display(frac(a, 24))$] for #math.equation(block: false, alt: "a")[$a$]. - Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis 64.5 degrees close parenthesis equals the fraction b over 24")[$cos ( 64.5^(∘) ) = display(frac(b, 24))$] for #math.equation(block: false, alt: "b")[$b$]. - Subtract #math.equation(block: false, alt: "64.5 degrees")[$64.5^(∘)$] from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] to find #math.equation(block: false, alt: "B")[$B$]. #math.equation(block: true, alt: "B equals 44 degrees , b equals 0.6")[$B = 44^(∘) , " " b = 0.6$] For Problems 17–22, find the labeled angle. Round your answer to tenths of a degree. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "74.2 degrees")[$74.2^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "56.4 degrees")[$56.4^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "66.0 degrees")[$66.0^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 23–28, evaluate the expression and sketch a right triangle to illustrate. #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis 0.2 close parenthesis")[$sin^(− 1) ( 0.2 )$] #math.equation(block: true, alt: "11.5 degrees")[$11.5^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis 0.8 close parenthesis")[$cos^(− 1) ( 0.8 )$] #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis 1.5 close parenthesis")[$tan^(− 1) ( 1.5 )$] #math.equation(block: true, alt: "56.3 degrees")[$56.3^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis 2.5 close parenthesis")[$tan^(− 1) ( 2.5 )$] #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis 0.2839 close parenthesis")[$cos^(− 1) ( 0.2839 )$] #math.equation(block: true, alt: "73.5 degrees")[$73.5^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis 0.4127 close parenthesis")[$sin^(− 1) ( 0.4127 )$] For Problems 29–32, write two different equations for the statement. The cosine of #math.equation(block: false, alt: "15 degrees")[$15^(∘)$] is #math.equation(block: false, alt: "0.9659")[$0.9659$]. #math.equation(block: false, alt: "cos ⁡ open parenthesis 15 degrees close parenthesis equals 0.9659")[$cos ( 15^(∘) ) = 0.9659 " "$] and #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis 0.9659 close parenthesis equals 15 degrees")[$" " cos^(− 1) ( 0.9659 ) = 15^(∘)$] The sine of #math.equation(block: false, alt: "70 degrees")[$70^(∘)$] is #math.equation(block: false, alt: "0.9397")[$0.9397$]. The angle whose tangent is #math.equation(block: false, alt: "3.1445")[$3.1445$] is #math.equation(block: false, alt: "65 degrees")[$65^(∘)$]. #math.equation(block: false, alt: "tan ⁡ open parenthesis 65 degrees close parenthesis equals 2.1445")[$tan ( 65^(∘) ) = 2.1445 " "$] and #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis 2.1445 close parenthesis equals 65 degrees")[$" " tan^(− 1) ( 2.1445 ) = 65^(∘)$] The angle whose cosine is #math.equation(block: false, alt: "0.0872")[$0.0872$] is #math.equation(block: false, alt: "85 degrees")[$85^(∘)$]. Evaluate the expressions, and explain what each means. #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis 0.6 close parenthesis , open parenthesis sin ⁡ 6 degrees close parenthesis to the power minus 1")[$sin^(− 1) ( 0.6 ) , " " " " " " ( sin 6^(∘) )^(− 1)$] #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis 0.6 close parenthesis approximately equals 36.87 degrees")[$sin^(− 1) ( 0.6 ) ≈ 36.87^(∘)$] is the angle whose sine is #math.equation(block: false, alt: "0.6")[$0.6$]. #math.equation(block: false, alt: "open parenthesis sin ⁡ 6 degrees close parenthesis to the power minus 1 approximately equals 9.5668")[$( sin 6^(∘) )^(− 1) ≈ 9.5668$] is the reciprocal of #math.equation(block: false, alt: "sin ⁡ open parenthesis 6 degrees close parenthesis")[$sin ( 6^(∘) )$]. Evaluate the expressions, and explain what each means. #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis 0.36 close parenthesis , open parenthesis cos ⁡ 36 degrees close parenthesis to the power minus 1")[$cos^(− 1) ( 0.36 ) , " " " " " " ( cos 36^(∘) )^(− 1)$] For Problems 35–38, + 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. The gondola cable for the ski lift at Snowy Peak is 2458 yards long and climbs 1860 feet. What angle with the horizontal does the cable make? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1806 over 3 open parenthesis 2458 close parenthesis , 14.6 degrees")[$sin ( θ ) = display(frac(1806, 3 ( 2458 ))) , " " 14.6^(∘)$] The Leaning Tower of Pisa is 55 meters in length. An object dropped from the top of the tower lands 4.8 meters from the base of the tower. At what angle from the horizontal does the tower lean? A mining company locates a vein of minerals at a depth of 32 meters. However, there is a layer of granite directly above the minerals, so they decide to drill at an angle, starting 10 meters from their original location. At what angle from the horizontal should they drill? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "tan ⁡ θ equals the fraction 32 over 10 , 72.6 degrees")[$tan θ = display(frac(32, 10)) , " " 72.6^(∘)$] The birdhouse in Carolyn's front yard is 12 feet tall, and its shadow at 4 pm is 15 feet 4 inches long. What is the angle of elevation of the sun at 4 pm? For Problems 39–42, + Sketch the right triangle described. + Solve the triangle. #math.equation(block: true, alt: "a equals 18 , b equals 26")[$a = 18 , " " b = 26$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "c equals 10 the square root of 10 approximately equals 31.6")[$c = 10 sqrt(10) ≈ 31.6$], #math.equation(block: false, alt: "A approximately equals 34.7 degrees")[$" " A ≈ 34.7^(∘)$], #math.equation(block: false, alt: "B approximately equals 55.3 degrees")[$" " B ≈ 55.3^(∘)$] #math.equation(block: true, alt: "a equals 35 , b equals 27")[$a = 35 , " " b = 27$] #math.equation(block: true, alt: "b equals 10.6 , c equals 19.2")[$b = 10.6 , " " c = 19.2$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "a equals the square root of 256.28 approximately equals 16.0")[$a = sqrt(256.28) ≈ 16.0$], #math.equation(block: false, alt: "A approximately equals 56.5 degrees")[$" " A ≈ 56.5^(∘)$], #math.equation(block: false, alt: "B approximately equals 33.5 degrees")[$" " B ≈ 33.5^(∘)$] #math.equation(block: true, alt: "a equals 88 , c equals 132")[$a = 88 , " " c = 132$] For Problems 43–48, + Make a sketch that illustrates the situation. Label your sketch with the given information. + Write an equation and solve the problem. The Mayan pyramid of El Castillo at Chichen Itza in Mexico has 91 steps. Each step is 26 cm high and 30 cm deep. + What angle does the side of the pyramid make with the horizontal? + What is the distance up the face of the pyramid, from base to top platform? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis the fraction 26 over 30 close parenthesis approximately equals 40.9 degrees , 91 the square root of 1676 approximately equals 3612.6")[$tan^(− 1) ( display(frac(26, 30)) ) ≈ 40.9^(∘) , " " " " 91 sqrt(1676) ≈ 3612.6$] cm An airplane begins its descent when its altitude is 10 kilometers. The angle of descent should be #math.equation(block: false, alt: "3 degrees")[$3^(∘)$] from horizontal. + How far from the airport (measured along the ground) should the airplane begin its descent? + How far will the airplane travel on its descent to the airport? A communications satellite is in a low earth orbit (LOE) at an altitude of 400 km. From the satellite, the angle of depression to earth's horizon is #math.equation(block: false, alt: "19.728 degrees")[$19.728^(∘)$]. Use this information to calculate the radius of the earth. + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "6415")[$6415$] km The first Ferris wheel was built for the 1893 Chicago world's fair. It had a diameter of 250 feet, and the boarding platform, at the base of the wheel, was 14 feet above the ground. If you boarded the wheel and rotated through an angle of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$], what would be your height above the ground? To find the distance across a ravine, Delbert takes some measurements from a small airplane. When he is a short distance from the ravine at an altitude of 500 feet, he finds that the angle of depression to the near side of the ravine is #math.equation(block: false, alt: "56 degrees")[$56^(∘)$], and the angle of depression to the far side is #math.equation(block: false, alt: "32 degrees")[$32^(∘)$]. What is the width of the ravine? (Hint: First find the horizontal distance from Delbert to the near side of the ravine.) + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "462.9")[$462.9$] ft The window in Francine's office is 4 feet wide and 5 feet tall. The bottom of the window is 3 feet from the floor. When the sun is at an angle of elevation of #math.equation(block: false, alt: "64 degrees")[$64^(∘)$], what is the area of the rectangular sunny spot on the floor? Assume that the edge of the sunny spot nearest the window is the same width as the window. Which of the following numbers are equal to #math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis")[$cos ( 45^(∘) )$]? + #math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$] + #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 2 over the square root of 2")[$display(frac(2, sqrt(2)))$] + #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$] (a) and (b) Which of the following numbers are equal to #math.equation(block: false, alt: "tan ⁡ open parenthesis 30 degrees close parenthesis")[$tan ( 30^(∘) )$]? + #math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$] + #math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$] + #math.equation(block: false, alt: "the fraction the square root of 3 over 3")[$display(frac(sqrt(3), 3))$] + #math.equation(block: false, alt: "the fraction 3 over the square root of 3")[$display(frac(3, sqrt(3)))$] Which of the following numbers are equal to #math.equation(block: false, alt: "tan ⁡ open parenthesis 60 degrees close parenthesis")[$tan ( 60^(∘) )$]? + #math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$] + #math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$] + #math.equation(block: false, alt: "the fraction the square root of 3 over 3")[$display(frac(sqrt(3), 3))$] + #math.equation(block: false, alt: "the fraction 3 over the square root of 3")[$display(frac(3, sqrt(3)))$] (a) and (d) Which of the following numbers are equal to #math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis")[$sin ( 60^(∘) )$]? + #math.equation(block: false, alt: "the fraction 3 over the square root of 2")[$display(frac(3, sqrt(2)))$] + #math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$] + #math.equation(block: false, alt: "the fraction the square root of 2 over 3")[$display(frac(sqrt(2), 3))$] + #math.equation(block: false, alt: "the fraction 2 over the square root of 3")[$display(frac(2, sqrt(3)))$] For Problems 53–58, choose all values from the list below that are exactly equal to, or decimal approximations for, the given trig ratio. (Try not to use a calculator!) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "sin ⁡ open parenthesis 30 degrees close parenthesis")[$sin ( 30^(∘) )$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 45 degrees close parenthesis")[$cos ( 45^(∘) )$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis 60 degrees close parenthesis")[$sin ( 60^(∘) )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis 45 degrees close parenthesis")[$tan ( 45^(∘) )$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis 60 degrees close parenthesis")[$tan ( 60^(∘) )$]]), [#math.equation(block: false, alt: "0.5000")[$" " " " " " 0.5000 " " " " " "$]], [#math.equation(block: false, alt: "0.5774")[$" " " " " " 0.5774 " " " " " "$]], [#math.equation(block: false, alt: "0.7071")[$" " " " " " 0.7071 " " " " " "$]], [#math.equation(block: false, alt: "0.8660")[$" " " " " " 0.8660 " " " " " "$]], [#math.equation(block: false, alt: "1.0000")[$" " " " " " 1.0000 " " " " " "$]], [#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 2 over the square root of 2")[$display(frac(2, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 3 over the square root of 2")[$display(frac(3, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$]], [#math.equation(block: false, alt: "the fraction 2 over the square root of 3")[$display(frac(2, sqrt(3)))$]], [#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 square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 3")[$display(frac(sqrt(3), 3))$]], )) #math.equation(block: true, alt: "cos ⁡ open parenthesis 30 degrees close parenthesis")[$cos ( 30^(∘) )$] #math.equation(block: true, alt: "the fraction the square root of 3 over 2 approximately equals 0.8660")[$display(frac(sqrt(3), 2)) ≈ 0.8660$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 45 degrees close parenthesis")[$sin ( 45^(∘) )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis 30 degrees close parenthesis")[$tan ( 30^(∘) )$] #math.equation(block: true, alt: "the fraction 1 over the square root of 3 equals the fraction the square root of 3 over 3 approximately equals 0.5774")[$display(frac(1, sqrt(3))) = display(frac(sqrt(3), 3)) ≈ 0.5774$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 60 degrees close parenthesis")[$cos ( 60^(∘) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 90 degrees close parenthesis")[$sin ( 90^(∘) )$] #math.equation(block: true, alt: "1.0000")[$1.0000$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 0 degrees close parenthesis")[$cos ( 0^(∘) )$] Fill in the table from memory with exact values. Do you notice any patterns that might help you learn the values? #figure(table( columns: 6, 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: "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")[$tan ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], )) #figure(table( columns: 6, 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: "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 the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#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: "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 the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#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")[$sqrt(3)$]], [undefined], )) Fill in the table from memory with decimal approximations to four places. #figure(table( columns: 6, 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: "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")[$tan ( θ )$]], [$" "$], [$" "$], [$" "$], [$" "$], [$" "$], )) For Problems 61 and 62, compare the given value with the trig ratios of the special angles to answer the questions. Try not to use a calculator. Is the acute angle larger or smaller than #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]? + #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis equals 0.7")[$sin ( α ) = 0.7$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis β close parenthesis equals 1.2")[$tan ( β ) = 1.2$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis γ close parenthesis equals 0.65")[$cos ( γ ) = 0.65$] + smaller + larger + larger Is the acute angle larger or smaller than #math.equation(block: false, alt: "60 degrees")[$60^(∘)$]? + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.75")[$cos ( θ ) = 0.75$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis ϕ close parenthesis equals 1.5")[$tan ( ϕ ) = 1.5$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis ψ close parenthesis equals 0.72")[$sin ( ψ ) = 0.72$] For Problems 63–72, solve the triangle. Give your answers as exact values. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "a equals 3 the square root of 3 , b equals 3 , B equals 30 degrees")[$a = 3 sqrt(3) , " " b = 3 , " " B = 30^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "a equals b equals 4 the square root of 2 , B equals 45 degrees")[$a = b = 4 sqrt(2) , " " B = 45^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "e equals 4 , f equals 4 the square root of 3 , F equals 120 degrees")[$e = 4 , " " f = 4 sqrt(3) , " " F = 120^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "d equals 2 the square root of 3 , e equals 2 the square root of 2 , f equals the square root of 2 plus the square root of 6 , F equals 75 degrees")[$d = 2 sqrt(3) , " " e = 2 sqrt(2) , f = sqrt(2) + sqrt(6) , " " F = 75^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "a equals 20 , b equals 20 , c equals 20 the square root of 2")[$a = 20 , " " b = 20 , " " c = 20 sqrt(2)$] #figure(figph[triangle], alt: "triangle", caption: none) + Find the perimeter of a regular hexagon if the apothegm is 8 cm long. (The #strong[apothegm] is the segment from the center of the hexagon and perpendicular to one of its sides.) + Find the area of the hexagon. #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "32 the square root of 3")[$32 sqrt(3)$] cm + #math.equation(block: false, alt: "128 the square root of 3")[$128 sqrt(3)$] sq cm Triangle #math.equation(block: false, alt: "A B C")[$A B C$] is equilateral, and its angle bisectors meet at point #math.equation(block: false, alt: "P")[$P$]. The sides of #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] are 6 inches long. Find the length of #math.equation(block: false, alt: "A P")[$A P$]. #figure(figph[triangle], alt: "triangle", caption: none) Find an exact value for the area of each triangle. + #figure(figph[triangle], alt: "triangle", caption: none) + #figure(figph[triangle], alt: "triangle", caption: none) + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "10")[$10$] sq cm + #math.equation(block: false, alt: "10 the square root of 2")[$10 sqrt(2)$] sq cm + #math.equation(block: false, alt: "10 the square root of 3")[$10 sqrt(3)$] sq cm Find an exact value for the perimeter of each parallelogram. + #figure(figph[triangle], alt: "triangle", caption: none) + #figure(figph[triangle], alt: "triangle", caption: none) + #figure(figph[triangle], alt: "triangle", caption: none) + Find the area of the outer square. + Find the dimensions and the area of the inner square. + What is the ratio of the area of the outer square to the area of the inner square? #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "64")[$64$] sq in + #math.equation(block: false, alt: "4 the square root of 2")[$4 sqrt(2)$] by #math.equation(block: false, alt: "4 the square root of 2")[$4 sqrt(2)$], area #math.equation(block: false, alt: "32")[$32$] sq in + #math.equation(block: false, alt: "2 : 1")[$2 : 1$] + Find the area of the inner square. + Find the dimensions and the area of the outer square. + What is the ratio of the area of the outer square to the area of the inner square? #figure(figph[triangle], alt: "triangle", caption: none)