#set document(title: "3.3 The Law of Cosines", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.3#h(0.6em)The Law of Cosines If we know two angles and one side of a triangle, we can use the Law of Sines to solve the triangle. We can also use the Law of Sines when we know two sides and the angle opposite one of them. But the Law of Sines is not helpful for the problem that opened this chapter, finding the distance from Avery to Clio. In this case we know two sides of the triangle, #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "c")[$c$], and the #emph[included] angle, #math.equation(block: false, alt: "B")[$B$]. #figure(figph[oblique triangles], alt: "oblique triangles", caption: none) To solve a triangle when we know two sides and the included angle, we will need a generalization of the Pythagorean theorem known as the Law of Cosines. In a right triangle, with #math.equation(block: false, alt: "C equals 90 degrees")[$C = 90^(∘)$], the Pythagorean theorem tells us that #math.equation(block: true, alt: "c squared equals a squared plus b squared")[$c^(2) = a^(2) + b^(2)$] If we allow angle #math.equation(block: false, alt: "C")[$C$] to vary, but keep #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] the same length, the side #math.equation(block: false, alt: "c")[$c$] will grow or shrink, depending on whether we increase or decrease the angle #math.equation(block: false, alt: "C")[$C$], as shown below. The sweep this sentence describes. Two sides are pinned: b = 4 lies along the horizontal axis and a = 3 reaches from the same corner at the angle C, its endpoint B riding the dashed circle of radius 3. The red third side c connects the two far endpoints, and the Law of Cosines c² = a² + b² − 2ab·cos(C) says exactly how its length responds: drag C larger and c stretches, drag it smaller and c shrinks. Stop at C = 90°, where cos(C) = 0 and the correction term vanishes: the law collapses to the Pythagorean theorem c² = a² + b², and with a = 3, b = 4 the red side is exactly 5 — the 3-4-5 right triangle sitting inside the general law as its special case.{"functions":\[{"color":"\#6a6a6a","expression":"3\*cos(t)","expression2":"3\*sin(t)","id":"b-locus","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*4","expression2":"0","id":"side-b","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*3\*cos(C\*pi/180)","expression2":"t\*3\*sin(C\*pi/180)","id":"side-a","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"4 + t\*(3\*cos(C\*pi/180) - 4)","expression2":"t\*3\*sin(C\*pi/180)","id":"side-c","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#2563eb","expression":"0.5\*cos(C\*pi/180\*t)","expression2":"0.5\*sin(C\*pi/180\*t)","id":"angle-arc","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"3\*cos(C\*pi/180) + 0.12\*cos(t)","expression2":"3\*sin(C\*pi/180) + 0.12\*sin(t)","id":"vertex-b","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-C","label":"Included angle C","max":160,"min":20,"name":"C","step":10,"unit":"\\u00b0","value":60}\],"title":"c\\u00b2 = a\\u00b2 + b\\u00b2 \\u2212 2ab\\u00b7cos(C) with a = 3, b = 4","version":1,"viewport":{"centerX":0.8,"centerY":1.1,"scale":55}}#figure(figph[three triangles], alt: "three triangles", caption: none) The Pythagorean theorem is actually a special case of a more general law that applies to all triangles, no matter what the size of angle #math.equation(block: false, alt: "C")[$C$]. The equation relating the three sides of a triangle is #math.equation(block: true, alt: "c squared equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis")[$c^(2) = a^(2) + b^(2) − 2 a b cos ( C )$] You can see that when #math.equation(block: false, alt: "C")[$C$] is a right angle, #math.equation(block: false, alt: "cos ⁡ 90 degrees equals 0")[$cos 90^(∘) = 0$], so the equation reduces to the Pythagorean theorem. We can write similar equations involving the angles or #math.equation(block: false, alt: "A")[$A$] or #math.equation(block: false, alt: "B")[$B$]. The three equations are all versions of the Law of Cosines. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Law of Cosines] If the angles of a triangle are #math.equation(block: false, alt: "A , B")[$A , B$], and #math.equation(block: false, alt: "C")[$C$], and the opposite sides are respectively #math.equation(block: false, alt: "a , b ,")[$a , b ,$] and #math.equation(block: false, alt: "c")[$c$], then #math.equation(block: true, alt: "a squared equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis; b squared equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis; c squared equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis")[$bold(italic(a))^(2) = bold(italic(b))^(2) + bold(italic(c))^(2) − 2 bold(italic(b)) bold(italic(c)) bold(cos) ( bold(italic(A)) ) \ bold(italic(b))^(2) = bold(italic(a))^(2) + bold(italic(c))^(2) − 2 bold(italic(a)) bold(italic(c)) bold(cos) ( bold(italic(B)) ) \ bold(italic(c))^(2) = bold(italic(a))^(2) + bold(italic(b))^(2) − 2 bold(italic(a)) bold(italic(b)) bold(cos) ( bold(italic(C)) )$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For a proof of the Law of Cosines, see Homework Problems 57 and 58. ] === Finding a Side Now we can solve the problem of the distance from Avery to Clio. Here is the figure from Section 3.1 showing the location of the three towns. #figure(figph[triangle between towns], alt: "triangle between towns", caption: none) #examplebox("Example 1")[][ How far is it from Avery to Clio? #solutionbox[ The angle #math.equation(block: false, alt: "∠ A B C equals 35 degrees plus 90 degrees equals 125 degrees")[$upright(∠) A B C = 35^(∘) + 90^(∘) = 125^(∘)$]. Thus, in #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] we have #math.equation(block: false, alt: "a equals 34 , c equals 48")[$a = 34 , " " c = 48$] and #math.equation(block: false, alt: "B equals 125 degrees")[$B = 125^(∘)$]. The distance from Avery to Clio is represented by #math.equation(block: false, alt: "b")[$b$] in the figure. #figure(figph[oblique triangle], alt: "oblique triangle", caption: none) We know two sides and the included angle, and we choose the version of the Law of Cosines that uses our known angle, #math.equation(block: false, alt: "B")[$B$], and substitute the other known values. #math.equation(block: true, alt: "b squared, equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis; b squared, equals 34 squared plus 48 squared minus 2 open parenthesis 34 close parenthesis open parenthesis 48 close parenthesis cos ⁡ open parenthesis 125 degrees close parenthesis, Simplify the right side.; b squared, equals 3460 minus 3264 cos ⁡ open parenthesis 125 degrees close parenthesis equals 5332.153, Take square roots.; b, equals 73.02")[$b^(2) & = a^(2) + c^(2) − 2 a c cos ( B ) \ b^(2) & = 34^(2) + 48^(2) − 2 ( 34 ) ( 48 ) cos ( 125^(∘) ) & & "Simplify the right side." \ b^(2) & = 3460 − 3264 cos ( 125^(∘) ) = 5332.153 & & "Take square roots." \ b & = 73.02$] Avery is about 73 miles from Clio. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ When simplifying the Law of Cosines, be careful to follow the order of operations. In the previous example, the right side of the equation #math.equation(block: true, alt: "b squared, equals 3460 minus 3264 cos ⁡ open parenthesis 125 degrees close parenthesis; b squared, equals 3460 minus 3264 open parenthesis minus 0.5736 close parenthesis")[$b^(2) & = 3460 − 3264 cos ( 125^(∘) ) \ b^(2) & = 3460 − 3264 ( − 0.5736 )$] has two terms, because 3264 is the coefficient of #math.equation(block: false, alt: "cos ⁡ open parenthesis 125 degrees close parenthesis")[$cos ( 125^(∘) )$], so it would be incorrect to subtract 3264 from 3460; we should multiply #math.equation(block: false, alt: "3264 open parenthesis minus 0.5736 close parenthesis")[$3264 ( − 0.5736 )$] first. If you are using a graphing calculator, you can enter the right side of the equation exactly as it is written. ] In #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], #math.equation(block: false, alt: "a equals 11 , c equals 23")[$a = 11 , " " c = 23$], and #math.equation(block: false, alt: "B equals 87 degrees")[$B = 87^(∘)$]. Find #math.equation(block: false, alt: "b")[$b$], and round your answer to two decimal places. We use the Law of Cosines. #math.equation(block: true, alt: "b squared equals 11 squared plus 23 squared minus 11 open parenthesis 23 close parenthesis cos ⁡ open parenthesis 87 degrees close parenthesis equals 623.52")[$b^(2) = 11^(2) + 23^(2) − 11 ( 23 ) cos ( 87^(∘) ) = 623.52$] so #math.equation(block: false, alt: "b equals the square root of 623.52 equals 24.97")[$b = sqrt(623.52) = 24.97$]. === Finding an Angle We can also use the Law of Cosines to find an angle when we know all three sides of a triangle. Pay close attention to the algebraic steps used to solve the equation in the next example. #examplebox("Example 2")[][ In the triangle at right, #math.equation(block: false, alt: "a equals 6 , b equals 7")[$a = 6 , " " b = 7$], and #math.equation(block: false, alt: "c equals 11")[$c = 11$]. Find angle #math.equation(block: false, alt: "C")[$C$]. #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We choose the version of the Law of Cosines that uses angle #math.equation(block: false, alt: "C")[$C$]. #math.equation(block: true, alt: "c squared, equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis, Substitute the known values.; 11 squared, equals 6 squared plus 7 squared minus 2 open parenthesis 6 close parenthesis open parenthesis 7 close parenthesis cos ⁡ open parenthesis C close parenthesis, Simplify each side.; 121, equals 36 plus 49 minus 84 cos ⁡ open parenthesis C close parenthesis, Isolate the cosine term.; 36, equals minus 84 cos ⁡ open parenthesis C close parenthesis, Solve for cos (C).; the fraction minus 3 over 7, equals cos ⁡ open parenthesis C close parenthesis, Solve for C.; C, equals cos to the power minus 1 ⁡ open parenthesis the fraction minus 3 over 7 close parenthesis equals 115.4 degrees")[$c^(2) & = a^(2) + b^(2) − 2 a b cos ( C ) & & "Substitute the known values." \ 11^(2) & = 6^(2) + 7^(2) − 2 ( 6 ) ( 7 ) cos ( C ) & & "Simplify each side." \ 121 & = 36 + 49 − 84 cos ( C ) & & "Isolate the cosine term." \ 36 & = − 84 cos ( C ) & & "Solve for cos (C)." \ display(frac(− 3, 7)) & = cos ( C ) & & "Solve for C." \ C & = cos^(− 1) ( display(frac(− 3, 7)) ) = 115.4^(∘)$] Angle #math.equation(block: false, alt: "C")[$C$] is about #math.equation(block: false, alt: "115.4 degrees")[$115.4^(∘)$]. ] ] In #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], #math.equation(block: false, alt: "a equals 5.3 , b equals 4.7")[$" " a = 5.3 , " " b = 4.7$], and #math.equation(block: false, alt: "c equals 6.1")[$" " c = 6.1$]. Find angle #math.equation(block: false, alt: "B")[$B$], and round your answer to two decimal places. Substitute the known values into the Law of Cosines to get #math.equation(block: true, alt: "4.7 squared equals 5.3 squared plus 6.1 squared minus 2 open parenthesis 5.3 close parenthesis open parenthesis 6.1 close parenthesis cos ⁡ open parenthesis B close parenthesis")[$4.7^(2) = 5.3^(2) + 6.1^(2) − 2 ( 5.3 ) ( 6.1 ) cos ( B )$] from which we find #math.equation(block: false, alt: "cos ⁡ open parenthesis B close parenthesis equals 0.6683")[$cos ( B ) = 0.6683$], so #math.equation(block: true, alt: "B equals cos to the power minus 1 ⁡ open parenthesis 0.6683 close parenthesis equals 48.07 degrees")[$B = cos^(− 1) ( 0.6683 ) = 48.07^(∘)$] . Once we have calculated one of the angles in a triangle, we can use either the Law of Sines or the Law of Cosines to find a second angle. Here is how we would use the Law of Sines to find angle #math.equation(block: false, alt: "A")[$A$] in the previous Example. #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis A close parenthesis over a, equals the fraction sin ⁡ open parenthesis C close parenthesis over c, Substitute the known values.; the fraction sin ⁡ open parenthesis A close parenthesis over 6, equals the fraction sin ⁡ open parenthesis 115.4 degrees close parenthesis over 11, Solve for sin ⁡ A .; sin ⁡ open parenthesis A close parenthesis, equals 6 times the fraction sin ⁡ open parenthesis 115.4 degrees close parenthesis over 11 approximately equals 0.4928")[$display(frac(sin ( A ), a)) & = display(frac(sin ( C ), c)) & & "Substitute the known values." \ display(frac(sin ( A ), 6)) & = display(frac(sin ( 115.4^(∘) ), 11)) & & "Solve for" bold(sin) bold(italic(A)) . \ sin ( A ) & = 6 ⋅ display(frac(sin ( 115.4^(∘) ), 11)) ≈ 0.4928$] Thus, #math.equation(block: false, alt: "A equals sin to the power minus 1 ⁡ open parenthesis 0.4928 close parenthesis equals 29.5 degrees")[$A = sin^(− 1) ( 0.4928 ) = 29.5^(∘)$]. There are two angles with sine #math.equation(block: false, alt: "29.5 degrees")[$29.5^(∘)$], but we know that #math.equation(block: false, alt: "A")[$A$] is an acute angle because it is opposite the shortest side of the triangle. Finally, #math.equation(block: true, alt: "B equals 180 degrees minus open parenthesis A plus C close parenthesis approximately equals 35.1 degrees")[$B = 180^(∘) − ( A + C ) ≈ 35.1^(∘)$] Alternatively, we can use the Law of Cosines to find angle #math.equation(block: false, alt: "A")[$A$]. #math.equation(block: true, alt: "a squared, equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis, Substitute the known values.; 6 squared, equals 7 squared plus 11 squared minus 2 open parenthesis 7 close parenthesis open parenthesis 11 close parenthesis cos ⁡ open parenthesis A close parenthesis, Simplify each side.; 36, equals 49 plus 121 minus 154 cos ⁡ open parenthesis A close parenthesis, Isolate the cosine term.; minus 134, equals minus 154 cos ⁡ open parenthesis A close parenthesis, Solve for cos ⁡ open parenthesis A close parenthesis .; the fraction 67 over 77, equals cos ⁡ open parenthesis A close parenthesis, Solve for A.; A, equals cos to the power minus 1 ⁡ open parenthesis the fraction 67 over 77 close parenthesis equals 29.5 degrees")[$a^(2) & = b^(2) + c^(2) − 2 b c cos ( A ) & & "Substitute the known values." \ 6^(2) & = 7^(2) + 11^(2) − 2 ( 7 ) ( 11 ) cos ( A ) & & "Simplify each side." \ 36 & = 49 + 121 − 154 cos ( A ) & & "Isolate the cosine term." \ − 134 & = − 154 cos ( A ) & & "Solve for" bold(cos) ( bold(italic(A)) ) . \ display(frac(67, 77)) & = cos ( A ) & & "Solve for A." \ A & = cos^(− 1) ( display(frac(67, 77)) ) = 29.5^(∘)$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ You may notice that the Law of Cosines requires more calculation than the Law of Sines, but there is only one possible angle with the given cosine. If we use the Law of Sines there are two possible angles with the given sine, and either or both may fit the triangle. ] === Navigation Even with the aid of GPS (Global Positioning System) instruments, aircraft pilots and ship captains need to understand navigation based on trigonometry. #examplebox("Example 3")[][ The sailing club leaves the marina on a heading #math.equation(block: false, alt: "15 degrees")[$15^(∘)$] east of north and sails for 18 miles. They then change course, and after traveling for 12 miles on a heading #math.equation(block: false, alt: "35 degrees")[$35^(∘)$] east of north, they experience engine trouble and radio for help. The marina sends a speed boat to rescue them. How far should the speed boat go, and on what heading? #solutionbox[ We'd like to find the distance #math.equation(block: false, alt: "b")[$b$] and the angle #math.equation(block: false, alt: "θ")[$θ$] shown in the figure. In #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], we can calculate the angle at point #math.equation(block: false, alt: "B")[$B$] where the sailing club changed course: #math.equation(block: true, alt: "B equals 180 degrees minus 35 degrees plus 15 degrees equals 160 degrees")[$B = 180^(∘) − 35^(∘) + 15^(∘) = 160^(∘)$] We know #math.equation(block: false, alt: "a equals 12")[$a = 12$] and #math.equation(block: false, alt: "c equals 18")[$c = 18$]. First we use the Law of Cosines to find the distance #math.equation(block: false, alt: "b")[$b$]. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "b squared, equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis, Substitute the known values.; equals 12 squared plus 18 squared minus 2 open parenthesis 12 close parenthesis 18 cos ⁡ open parenthesis 160 degrees close parenthesis, Evaluate.; equals 873.95, Take positive square root.; b, equals 29.56")[$b^(2) & = a^(2) + c^(2) − 2 a c cos ( B ) & & "Substitute the known values." \ & = 12^(2) + 18^(2) − 2 ( 12 ) 18 cos ( 160^(∘) ) & & "Evaluate." \ & = 873.95 & & "Take positive square root." \ b & = 29.56$] Next, we apply the Law of Cosines again to find #math.equation(block: false, alt: "∠ A")[$upright(∠) A$]. #math.equation(block: true, alt: "a squared, equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis; 12 squared, equals 29.56 squared plus 18 squared minus 2 open parenthesis 29.56 close parenthesis open parenthesis 18 close parenthesis cos ⁡ open parenthesis A close parenthesis, Solve for cos ⁡ open parenthesis A close parenthesis .; cos ⁡ open parenthesis A close parenthesis, equals the fraction 12 squared minus 29.56 squared minus 18 squared over minus 2 open parenthesis 29.56 close parenthesis open parenthesis 18 close parenthesis equals 0.9903; C, equals cos to the power minus 1 ⁡ open parenthesis 0.9903 close parenthesis equals 8 degrees")[$a^(2) & = b^(2) + c^(2) − 2 b c cos ( A ) \ 12^(2) & = 29.56^(2) + 18^(2) − 2 ( 29.56 ) ( 18 ) cos ( A ) & & "Solve for" bold(cos) ( bold(italic(A)) ) . \ cos ( A ) & = display(frac(12^(2) − 29.56^(2) − 18^(2), − 2 ( 29.56 ) ( 18 ))) = 0.9903 \ C & = cos^(− 1) ( 0.9903 ) = 8^(∘)$] Thus, #math.equation(block: false, alt: "b equals 29.56")[$b = 29.56$] and #math.equation(block: false, alt: "θ equals 8 degrees plus 15 degrees equals 23 degrees")[$" " θ = 8^(∘) + 15^(∘) = 23^(∘)$]. The speed boat should travel 29.56 miles on a heading #math.equation(block: false, alt: "23 degrees")[$23^(∘)$] east of north. ] ] Howard wants to fly from Anchorage to Nome, Alaska, a distance of 540 miles on a heading #math.equation(block: false, alt: "57 degrees")[$57^(∘)$] west of north. After flying for some time, he discovers that his heading is in error, and he is actually flying #math.equation(block: false, alt: "47 degrees")[$47^(∘)$] west of north. Howard corrects his flight plan and changes course when he is exactly 200 miles from Anchorage. What is his new heading, and how far is he from Nome? #figure(figph[triangle], alt: "triangle", caption: none) We use the Law of Cosines to find his distance from Nome: #math.equation(block: true, alt: "d squared equals 540 squared plus 200 squared minus 2 open parenthesis 200 close parenthesis open parenthesis 540 close parenthesis cos ⁡ open parenthesis 10 degrees close parenthesis")[$d^(2) = 540^(2) + 200^(2) − 2 ( 200 ) ( 540 ) cos ( 10^(∘) )$] Solving for #math.equation(block: false, alt: "d")[$d$] gives 344.8. Next, use either law to find that angle #math.equation(block: false, alt: "B")[$B$] is #math.equation(block: false, alt: "164.22 degrees")[$164.22^(∘)$]. The new heading is #math.equation(block: false, alt: "47 degrees")[$47^(∘)$] plus the supplement of #math.equation(block: false, alt: "B")[$B$], or #math.equation(block: false, alt: "47 degrees plus 15.78 degrees equals 62.78 degrees")[$47^(∘) + 15.78^(∘) = 62.78^(∘)$] west of north, and the distance is 344.8 miles. === Which Law to Use How can we decide which law, the Law of Sines or the Law of Cosines, is appropriate for a given problem? - If we are solving a #emph[right] triangle, we don't need the Laws of Sines and Cosines; all we need are the definitions of the trigonometric ratios. - But for oblique triangles, we can identify the following cases: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[How to Solve an Oblique Triangle] #figure(table( columns: 2, align: left, inset: 6pt, table.header([If we know:], [We can use:]), [1. One side and two angles (SAA)], [1. Law of Sines, to find another side], [2. Two sides and the included angle #linebreak() (SAS) #linebreak()], [2. Law of Cosines, to find the third #linebreak() side #linebreak()], [3. Three sides (SSS)], [3. Law of Cosines, to find an angle], [4. Two sides and the angle opposite #linebreak() one of them (SSA, the ambiguous #linebreak() case) #linebreak()], [4. Law of Sines, to find another angle, #linebreak() or Law of Cosines, to find another #linebreak() side #linebreak()], )) ] #examplebox("Example 4")[][ In the triangle at right, which law should you use to find #math.equation(block: false, alt: "∠ B")[$upright(∠) B$]? #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We know two sides of the triangle and the angle opposite one of them. We can use the Law of Sines to find #math.equation(block: false, alt: "∠ B")[$upright(∠) B$]. #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis B close parenthesis over 6, equals the fraction sin ⁡ open parenthesis 40 degrees close parenthesis over 10; sin ⁡ open parenthesis B close parenthesis, equals 0.3857")[$display(frac(sin ( B ), 6)) & = display(frac(sin ( 40^(∘) ), 10)) \ sin ( B ) & = 0.3857$] There are two angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] with sine #math.equation(block: false, alt: "0.3857")[$0.3857$], namely #math.equation(block: false, alt: "22.7 degrees")[$22.7^(∘)$] and its supplement, #math.equation(block: false, alt: "157.3 degrees")[$157.3^(∘)$]. Both of these angles might produce a solution. However, in this case we notice that because #math.equation(block: false, alt: "a greater than b")[$a > b$], angle #math.equation(block: false, alt: "B")[$B$] must be acute. ] ] In the triangle at right, which part of the triangle can you find, and which law should you use? #figure(figph[triangle], alt: "triangle", caption: none) We know two sides and the included angle, so we can find side #math.equation(block: false, alt: "c")[$c$] using the Law of Cosines. In the previous Example we used the Law of Sines to find an angle. Because there are always two angles with a given sine, we should check whether both possible angles result in a triangle. But here is another approach: We can apply the Law of Cosines to find the third side first. With that method, we'll need the quadratic formula. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quadratic Formula] The solutions of the quadratic equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 , a not equal to 0 ,")[$a x^(2) + b x + c = 0 , " " " " a ≠ 0 ,$] are given by #math.equation(block: true, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a")[$bold(italic(x)) = display(frac(− bold(italic(b)) ± sqrt(bold(italic(b))^(2) − 4 bold(italic(a)) bold(italic(c))), 2 bold(italic(a))))$] ] #examplebox("Example 5")[][ In the triangle at right, use the Law of Cosines to find #math.equation(block: false, alt: "c")[$c$]. #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We will use the Law of Cosines with angle #math.equation(block: false, alt: "A")[$A$]. #math.equation(block: true, alt: "a squared, equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis; 10 squared, equals 6 squared plus c squared minus 2 open parenthesis 6 close parenthesis c cos ⁡ open parenthesis 40 degrees close parenthesis; 100, equals 36 plus c squared minus 12 open parenthesis 0.7660 close parenthesis c")[$a^(2) & = b^(2) + c^(2) − 2 b c cos ( A ) \ 10^(2) & = 6^(2) + c^(2) − 2 ( 6 ) c cos ( 40^(∘) ) \ 100 & = 36 + c^(2) − 12 ( 0.7660 ) c$] This is a quadratic equation in #math.equation(block: false, alt: "c")[$c$], so we rearrange the terms into standard form. #math.equation(block: true, alt: "c squared minus 12 open parenthesis 0.7660 close parenthesis c minus 64, equals 0; c squared minus 9.1925 c minus 64, equals 0")[$c^(2) − 12 ( 0.7660 ) c − 64 & = 0 \ c^(2) − 9.1925 c − 64 & = 0$] Now we can solve for #math.equation(block: false, alt: "c")[$c$] using the quadratic formula. #math.equation(block: true, alt: "c equals the fraction 9.1925 plus or minus the square root of 9.1925 squared minus 4 open parenthesis 1 close parenthesis open parenthesis minus 64 close parenthesis over 2 open parenthesis 1 close parenthesis")[$c = display(frac(9.1925 ± sqrt(9.1925^(2) − 4 ( 1 ) ( − 64 )), 2 ( 1 )))$] There are two solutions, #math.equation(block: false, alt: "13.8")[$13.8$] and #math.equation(block: false, alt: "minus 4.6")[$− 4.6$]. We can ignore the negative solution, so #math.equation(block: false, alt: "c equals 13.8")[$c = 13.8$]. ] ] === Using the Law of Cosines for the Ambiguous Case In Section 3.2 we encountered the ambiguous case: If we know two sides #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] of a triangle and the acute angle #math.equation(block: false, alt: "α")[$α$] opposite one of them, there may be one solution, two solutions, or no solution, depending on the size of #math.equation(block: false, alt: "a")[$a$] in relation to #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "α")[$α$], as shown below. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Ambiguous Case] + No solution: #math.equation(block: false, alt: "a less than b sin ⁡ open parenthesis α close parenthesis")[$a < b sin ( α )$]#figure(figph[triangle], alt: "triangle", caption: none) #linebreak() #math.equation(block: false, alt: "α")[$α$] is too short to make a triangle. + One solution: #math.equation(block: false, alt: "a equals b sin ⁡ open parenthesis α close parenthesis")[$a = b sin ( α )$] #figure(figph[triangle], alt: "triangle", caption: none) #linebreak() #math.equation(block: false, alt: "α")[$α$] is exactly the right length to make a right triangle. + Two solutions: #math.equation(block: false, alt: "b sin ⁡ open parenthesis α close parenthesis less than a less than b")[$b sin ( α ) < a < b$]#figure(figph[triangle], alt: "triangle", caption: none) + One solution: #math.equation(block: false, alt: "a greater than b")[$a > b$] #figure(figph[triangle], alt: "triangle", caption: none) ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If #math.equation(block: false, alt: "α")[$α$] is an obtuse angle, things are simpler: there is one solution if #math.equation(block: false, alt: "a greater than b")[$a > b$], and no solution if #math.equation(block: false, alt: "a less than or equal to b")[$a ≤ b$]. ] To discover which case applies (no triangle, one triangle, or two triangles), we can begin by using the Law of Cosines to find the third side. As we saw in the previous Example, for this method we need the quadratic formula. A quadratic equation can have one solution, two solutions, or no solution, depending on the value of the #strong[discriminant], #math.equation(block: false, alt: "D equals b squared minus 4 a c")[$D = b^(2) − 4 a c$]. (Note that the discriminant is the expression under the radical in the quadratic formula. Perhaps you recall how the discriminant determines the number of solutions, and why.) - If #math.equation(block: false, alt: "D greater than 0")[$" " D > 0$], the equation has two solutions. - If #math.equation(block: false, alt: "D equals 0")[$" " D = 0$], the equation has one solution. - If #math.equation(block: false, alt: "D less than 0")[$" " D < 0$], the equation has no (real) solution. And the number of solutions determines the number of triangles that fit the given properties. - If the quadratic equation has one positive solution, there is one triangle. - If the quadratic equation has two positive solutions, there are two triangles. - If the quadratic equation has no positive solutions, there is no triangle with the given properties. #examplebox("Example 6")[][ In #math.equation(block: false, alt: "△ A B C , B equals 14.4 degrees , a equals 8")[$upright(△) A B C , " " B = 14.4^(∘) , " " a = 8$], and #math.equation(block: false, alt: "b equals 3")[$b = 3$]. Solve the triangle. #solutionbox[ We begin by finding the third side of the triangle. Using the Law of Cosines, we have #math.equation(block: true, alt: "b squared, equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis, Substitute the known values.; 3 squared, equals 8 squared plus c squared minus 2 open parenthesis 8 close parenthesis c cos ⁡ open parenthesis 14.4 degrees close parenthesis, Simplify.; 9, equals 64 plus c squared minus 16 c open parenthesis 0.9686 close parenthesis, Write the equation in standard form.; 0, equals c squared minus 15.497 c plus 55, Apply the quadratic formula.")[$b^(2) & = a^(2) + c^(2) − 2 a c cos ( B ) & & "Substitute the known values." \ 3^(2) & = 8^(2) + c^(2) − 2 ( 8 ) c cos ( 14.4^(∘) ) & & "Simplify." \ 9 & = 64 + c^(2) − 16 c ( 0.9686 ) & & "Write the equation in standard form." \ 0 & = c^(2) − 15.497 c + 55 & & "Apply the quadratic formula."$] #math.equation(block: true, alt: "c, equals the fraction 15.497 plus or minus the square root of open parenthesis minus 15.497 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis 55 close parenthesis over 2 open parenthesis 1 close parenthesis, Simplify.; equals the fraction 15.497 plus or minus 4.490 over 2 equals 5.503 or 9.994")[$c & = display(frac(15.497 ± sqrt(( − 15.497 )^(2) − 4 ( 1 ) ( 55 )), 2 ( 1 ))) & & "Simplify." \ & = display(frac(15.497 ± 4.490, 2)) = 5.503 " " " or " " " 9.994$] Because there are two positive solutions for side #math.equation(block: false, alt: "c")[$c$], either #math.equation(block: false, alt: "c equals 5.503")[$c = 5.503$] or #math.equation(block: false, alt: "c equals 9.994")[$c = 9.994$], there are two triangles with the given properties. We apply the Law of Cosines again to find angle #math.equation(block: false, alt: "C")[$C$] in each triangle. For the triangle with #math.equation(block: false, alt: "c equals 5.503")[$c = 5.503$], we have #math.equation(block: true, alt: "c squared, equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis, Substitute the known values.; 5.503 squared, equals 8 squared plus 3 squared minus 2 open parenthesis 8 close parenthesis open parenthesis 3 close parenthesis cos ⁡ open parenthesis C close parenthesis, Solve for cos ⁡ open parenthesis C close parenthesis .; cos ⁡ open parenthesis C close parenthesis, equals the fraction 5.503 squared minus 64 minus 9 over minus 48 equals 0.8899; C, equals cos to the power minus 1 ⁡ open parenthesis 0.8899 close parenthesis equals 27.1 degrees")[$c^(2) & = a^(2) + b^(2) − 2 a b cos ( C ) & & "Substitute the known values." \ 5.503^(2) & = 8^(2) + 3^(2) − 2 ( 8 ) ( 3 ) cos ( C ) & & "Solve for" bold(cos) ( bold(italic(C)) ) . \ cos ( C ) & = display(frac(5.503^(2) − 64 − 9, − 48)) = 0.8899 \ C & = cos^(− 1) ( 0.8899 ) = 27.1^(∘)$] and #math.equation(block: false, alt: "A equals 180 degrees minus open parenthesis 14.4 degrees plus 27.1 degrees close parenthesis equals 138.5 degrees")[$A = 180^(∘) − ( 14.4^(∘) + 27.1^(∘) ) = 138.5^(∘)$]. For the triangle with #math.equation(block: false, alt: "c equals 9.994")[$c = 9.994$], we have #math.equation(block: true, alt: "c squared, equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis, Substitute the known values.; 9.994 squared, equals 8 squared plus 3 squared minus 2 open parenthesis 8 close parenthesis open parenthesis 3 close parenthesis cos ⁡ open parenthesis C close parenthesis, Solve for cos ⁡ open parenthesis C close parenthesis .; cos ⁡ open parenthesis C close parenthesis, equals the fraction 9.994 squared minus 64 minus 9 over minus 48 equals minus 0.5600; C, equals cos to the power minus 1 ⁡ open parenthesis minus 0.5600 close parenthesis equals 124.1 degrees")[$c^(2) & = a^(2) + b^(2) − 2 a b cos ( C ) & & "Substitute the known values." \ 9.994^(2) & = 8^(2) + 3^(2) − 2 ( 8 ) ( 3 ) cos ( C ) & & "Solve for" bold(cos) ( bold(italic(C)) ) . \ cos ( C ) & = display(frac(9.994^(2) − 64 − 9, − 48)) = − 0.5600 \ C & = cos^(− 1) ( − 0.5600 ) = 124.1^(∘)$] and #math.equation(block: false, alt: "A equals 180 degrees minus open parenthesis 14.4 degrees plus 124.1 degrees close parenthesis equals 41.5 degrees")[$A = 180^(∘) − ( 14.4^(∘) + 124.1^(∘) ) = 41.5^(∘)$]. Both triangles are shown below. #figure(figph[two triangles], alt: "two triangles", caption: none) ] ] Use the Law of Cosines to find all triangles #math.equation(block: false, alt: "A B C")[$A B C$] with #math.equation(block: false, alt: "A equals 48 degrees , a equals 10")[$A = 48^(∘) , " " a = 10$], and #math.equation(block: false, alt: "b equals 15")[$b = 15$]. Using the Law of Cosines to find side #math.equation(block: false, alt: "c")[$c$], we get the equation #math.equation(block: true, alt: "a squared, equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis; 100, equals 225 plus c squared minus 30 c cos ⁡ open parenthesis 48 degrees close parenthesis")[$a^(2) & = b^(2) + c^(2) − 2 b c cos ( A ) \ 100 & = 225 + c^(2) − 30 c cos ( 48^(∘) )$] Putting this quadratic equation into standard form we get #math.equation(block: true, alt: "c squared minus 20.07 c plus 125 equals 0")[$c^(2) − 20.07 c + 125 = 0$] The discriminant is #math.equation(block: true, alt: "b squared minus 4 a c equals 20.07 squared minus 4 open parenthesis 125 close parenthesis equals minus 97.2")[$b^(2) − 4 a c = 20.07^(2) − 4 ( 125 ) = − 97.2$] Because the discriminant is negative, the equation has no solution, and there is no triangle satisfying the stated conditions. Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve each quadratic equation. + #math.equation(block: false, alt: "2.5 x squared plus 6.2 equals 166.2")[$2.5 x^(2) + 6.2 = 166.2$] + #math.equation(block: false, alt: "0.8 x squared minus 124 equals 376")[$0.8 x^(2) − 124 = 376$] + #math.equation(block: false, alt: "2 x open parenthesis 2 x minus 3 close parenthesis equals 208")[$2 x ( 2 x − 3 ) = 208$] + #math.equation(block: false, alt: "3 x open parenthesis x plus 5 close parenthesis equals 900")[$3 x ( x + 5 ) = 900$] + #math.equation(block: false, alt: "2 x squared minus 6 x equals 233.12")[$2 x^(2) − 6 x = 233.12$] + #math.equation(block: false, alt: "0.5 x squared plus 1.5 x equals 464")[$0.5 x^(2) + 1.5 x = 464$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "plus or minus 8")[$± 8$] + #math.equation(block: false, alt: "plus or minus 25")[$± 25$] + #math.equation(block: false, alt: "8 , minus 6.5")[$8 , " " − 6.5$] + #math.equation(block: false, alt: "15 , minus 20")[$15 , " " − 20$] + #math.equation(block: false, alt: "12.4 , minus 9.4")[$12.4 , " " − 9.4$] + #math.equation(block: false, alt: "29 , minus 32")[$29 , " " − 32$] ] === Section 3.3 Summary ==== Vocabulary - Quadratic equation - Quadratic formula - Discriminant ==== Concepts + The Law of Sines is not helpful when we know two sides of the triangle and the included angle. In this case we need the Law of Cosines. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Law of Cosines] If the angles of a triangle are #math.equation(block: false, alt: "A , B")[$A , B$], and #math.equation(block: false, alt: "C")[$C$], and the opposite sides are respectively #math.equation(block: false, alt: "a , b ,")[$a , b ,$] and #math.equation(block: false, alt: "c")[$c$], then #math.equation(block: true, alt: "a squared equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis; b squared equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis; c squared equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis")[$a^(2) = b^(2) + c^(2) − 2 b c cos ( A ) \ b^(2) = a^(2) + c^(2) − 2 a c cos ( B ) \ c^(2) = a^(2) + b^(2) − 2 a b cos ( C )$] ] + We can also use the Law of Cosines to find an angle when we know all three sides of a triangle. + We can use the Law of Cosines to solve the ambiguous case. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[How to Solve an Oblique Triangle] #figure(table( columns: 2, align: left, inset: 6pt, table.header([If we know:], [We can use:]), [1. One side and two angles (SAA)], [1. Law of Sines, to find another side], [2. Two sides and the angle opposite #linebreak() one of them (SSA, the ambiguous #linebreak() case) #linebreak()], [2. Law of Sines, to find another angle, #linebreak() or Law of Cosines, to find another #linebreak() side #linebreak()], [3. Two sides and the included angle #linebreak() (SAS) #linebreak()], [3. Law of Cosines, to find the third #linebreak() side #linebreak()], [4. Three sides (SSS)], [4. Law of Cosines, to find an angle], )) ] ==== Study Questions + The Law of Cosines is really a generalization of what familiar theorem? + If you know all three sides of a triangle and one angle, what might be the advantage in using the Law of Cosines o find another angle, instead of the Law of Sines? + State the quadratic formula from memory. Try to sing the quadratic formula to the tune of "Pop Goes the Weasel." + Francine is solving a triangle in which #math.equation(block: false, alt: "a equals 20 , b equals 16")[$a = 20 , " " b = 16$], and #math.equation(block: false, alt: "A equals 26 degrees")[$A = 26^(∘)$]. She finds that #math.equation(block: false, alt: "sin ⁡ B equals 0.3507")[$sin B = 0.3507$]. How does she know that #math.equation(block: false, alt: "B equals 20.5 degrees")[$B = 20.5^(∘)$], and not #math.equation(block: false, alt: "159.5 degrees")[$159.5^(∘)$]? ==== Skills + Use the Law of Cosines to find the side opposite an angle \#7-12 + Use the Law of Cosines to find an angle \#13-20 + Use the Law of Cosines to find a side adjacent to an angle \#21-26 + Decide which law to use \#27-34 + Solve a triangle \#35-42 + Solve problems using the Law of Cosines \#43-56 === Homework 3.3 + Simplify #math.equation(block: false, alt: "5 squared plus 7 squared minus 2 open parenthesis 5 close parenthesis open parenthesis 7 close parenthesis cos ⁡ open parenthesis θ close parenthesis")[$" " " " 5^(2) + 7^(2) − 2 ( 5 ) ( 7 ) cos ( θ )$] + Evaluate the expression in part (a) for #math.equation(block: false, alt: "θ equals 29 degrees")[$θ = 29^(∘)$] + Evaluate the expression in part (a) for #math.equation(block: false, alt: "θ equals 151 degrees")[$θ = 151^(∘)$] + #math.equation(block: false, alt: "74 minus 70 cos ⁡ open parenthesis θ close parenthesis")[$74 − 70 cos ( θ )$] + 12.78 + 135.22 + Simplify #math.equation(block: false, alt: "26.1 squared plus 32.5 squared minus 2 open parenthesis 26.1 close parenthesis open parenthesis 32.5 close parenthesis cos ⁡ open parenthesis ϕ close parenthesis")[$" " " " 26.1^(2) + 32.5^(2) − 2 ( 26.1 ) ( 32.5 ) cos ( ϕ )$] + Evaluate the expression in part (a) for #math.equation(block: false, alt: "ϕ equals 64 degrees")[$ϕ = 64^(∘)$] + Evaluate the expression in part (a) for #math.equation(block: false, alt: "ϕ equals 116 degrees")[$ϕ = 116^(∘)$] + Solve #math.equation(block: false, alt: "b squared equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis β close parenthesis")[$" " " " b^(2) = a^(2) + c^(2) − 2 a c cos ( β ) " " " "$] for #math.equation(block: false, alt: "cos ⁡ open parenthesis β close parenthesis")[$" " " " cos ( β )$] + For the equation in part (a), find #math.equation(block: false, alt: "cos ⁡ β")[$cos β$] if #math.equation(block: false, alt: "a equals 5 , b equals 11")[$a = 5 , " " b = 11$], and #math.equation(block: false, alt: "c equals 8")[$c = 8$]. + #math.equation(block: false, alt: "the fraction a squared plus c squared minus b squared over 2 a c")[$display(frac(a^(2) + c^(2) − b^(2), 2 a c))$] + #math.equation(block: false, alt: "minus 0.4")[$− 0.4$] + Solve #math.equation(block: false, alt: "a squared equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis α close parenthesis")[$" " " " a^(2) = b^(2) + c^(2) − 2 b c cos ( α ) " " " "$] for #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$" " " " cos ( α )$] + For the equation in part (a), find #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$cos ( α )$] if #math.equation(block: false, alt: "a equals 4.6 , b equals 7.2")[$a = 4.6 , " " b = 7.2$], and #math.equation(block: false, alt: "c equals 9.4")[$c = 9.4$]. + The equation #math.equation(block: false, alt: "9 squared equals b squared plus 4 squared minus 2 b open parenthesis 4 close parenthesis cos ⁡ open parenthesis α close parenthesis")[$" " " " 9^(2) = b^(2) + 4^(2) − 2 b ( 4 ) cos ( α ) " " " "$] is quadratic in #math.equation(block: false, alt: "b")[$b$]. Write the equation in standard form. + Solve the equation in part (a) for #math.equation(block: false, alt: "b")[$b$] if #math.equation(block: false, alt: "α equals 48 degrees")[$α = 48^(∘)$]. + #math.equation(block: false, alt: "b squared minus 8 cos ⁡ open parenthesis α close parenthesis b minus 65 equals 0")[$b^(2) − 8 cos ( α ) b − 65 = 0$] + #math.equation(block: false, alt: "11.17 , minus 5.82")[$11.17 , " " − 5.82$] + The equation #math.equation(block: false, alt: "6 squared equals 5 squared plus c squared minus 2 open parenthesis 5 close parenthesis c cos ⁡ open parenthesis β close parenthesis")[$" " " " 6^(2) = 5^(2) + c^(2) − 2 ( 5 ) c cos ( β ) " " " "$] is quadratic in #math.equation(block: false, alt: "c")[$c$]. Write the equation in standard form. + Solve the equation in part (a) for #math.equation(block: false, alt: "c")[$c$] if #math.equation(block: false, alt: "β equals 126 degrees")[$β = 126^(∘)$]. For Problems 7–12, use the Law of Cosines to find the indicated side. Round to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) 7.70 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 13.44 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 5.12 #figure(figph[triangle], alt: "triangle", caption: none) For Problems 13–16,use the Law of Cosines to find the indicated angle. Round to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "133.43 degrees")[$133.43^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "40.64 degrees")[$40.64^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 17–20, find the angles of the triangle. Round answers to two decimal places. #math.equation(block: true, alt: "a equals 23 , b equals 14 , c equals 18")[$a = 23 , " " b = 14 , " " c = 18$] #math.equation(block: true, alt: "A equals 91.02 degrees , B equals 37.49 degrees , C equals 51.49 degrees")[$A = 91.02^(∘) , " " B = 37.49^(∘) , " " C = 51.49^(∘)$] #math.equation(block: true, alt: "a equals 18 , b equals 25 , c equals 19")[$a = 18 , " " b = 25 , " " c = 19$] #math.equation(block: true, alt: "a equals 16.3 , b equals 28.1 , c equals 19.4")[$a = 16.3 , " " b = 28.1 , " " c = 19.4$] #math.equation(block: true, alt: "A equals 34.34 degrees , B equals 103.49 degrees , C equals 42.17 degrees")[$A = 34.34^(∘) , " " B = 103.49^(∘) , " " C = 42.17^(∘)$] #math.equation(block: true, alt: "a equals 82.3 , b equals 22.5 , c equals 66.8")[$a = 82.3 , " " b = 22.5 , " " c = 66.8$] For Problems 21–26, use the Law of Cosines to find the unknown side. Round your answers to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) 6.30 or 2.70 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 29.76 or 5.91 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 16.00 #figure(figph[triangle], alt: "triangle", caption: none) For Problems 27–34, which law should you use to find the labeled unknown value, the Law of Sines or the Law of Cosines? Write an equation you can solve to find the unknown value. For Problems 31-34, you may need two steps to find the unknown value. #figure(figph[triangle], alt: "triangle", caption: none) Law of Cosines: #math.equation(block: false, alt: "61 squared equals 29 squared plus 46 squared minus 2 times 29 times 46 cos ⁡ open parenthesis ϕ close parenthesis")[$61^(2) = 29^(2) + 46^(2) − 2 ⋅ 29 ⋅ 46 cos ( ϕ )$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) Law of Sines: #math.equation(block: false, alt: "the fraction a over sin ⁡ open parenthesis 46 degrees close parenthesis equals the fraction 16 over sin ⁡ open parenthesis 25 degrees close parenthesis")[$display(frac(a, sin ( 46^(∘) ))) = display(frac(16, sin ( 25^(∘) )))$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) First the Law of Cosines: #math.equation(block: false, alt: "x squared equals 47 squared plus 29 squared minus 2 times 47 times 29 cos ⁡ open parenthesis 81 degrees close parenthesis")[$x^(2) = 47^(2) + 29^(2) − 2 ⋅ 47 ⋅ 29 cos ( 81^(∘) )$], then either the Law of Sines: #math.equation(block: false, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over 47 equals the fraction sin ⁡ open parenthesis 81 degrees close parenthesis over x")[$display(frac(sin ( θ ), 47)) = display(frac(sin ( 81^(∘) ), x))$] or the Law of Cosines: #math.equation(block: false, alt: "47 squared equals x squared plus 29 squared minus 2 times x times 29 cos ⁡ open parenthesis θ close parenthesis")[$47^(2) = x^(2) + 29^(2) − 2 ⋅ x ⋅ 29 cos ( θ )$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) Law of Cosines: #math.equation(block: false, alt: "9 squared equals 4 squared plus z squared minus 2 times 4 times z cos ⁡ open parenthesis 28 degrees close parenthesis")[$9^(2) = 4^(2) + z^(2) − 2 ⋅ 4 ⋅ z cos ( 28^(∘) )$], or use the Law of Sines first to find the (acute) angle opposite the side of length 4, then find the angle opposite the side of length #math.equation(block: false, alt: "z")[$z$] by subtracting the sum of the known angles from #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], then using the Law of Sines again. #figure(figph[triangle], alt: "triangle", caption: none) For Problems 35–42, + Sketch and label the triangle. + Solve the triangle. Round answers to two decimal places. #math.equation(block: true, alt: "B equals 47 degrees , a equals 23 , c equals 17")[$B = 47^(∘) , " " a = 23 , " " c = 17$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "b equals 16.87")[$b = 16.87$], #math.equation(block: false, alt: "A equals 85.53 degrees")[$" " A = 85.53^(∘)$], #math.equation(block: false, alt: "C equals 47.47 degrees")[$" " C = 47.47^(∘)$] #math.equation(block: true, alt: "C equals 32 degrees , a equals 14 , b equals 18")[$C = 32^(∘) , " " a = 14 , " " b = 18$] #math.equation(block: true, alt: "a equals 8 , b equals 7 , c equals 9")[$a = 8 , " " b = 7 , " " c = 9$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "A equals 58.41 degrees")[$A = 58.41^(∘)$], #math.equation(block: false, alt: "B equals 48.19 degrees")[$B = 48.19^(∘)$], #math.equation(block: false, alt: "C equals 73.40 degrees")[$C = 73.40^(∘)$] #math.equation(block: true, alt: "a equals 23 , b equals 34 , c equals 45")[$a = 23 , " " b = 34 , " " c = 45$] #math.equation(block: true, alt: "b equals 72 , c equals 98 , B equals 38 degrees")[$b = 72 , " " c = 98 , " " B = 38^(∘)$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "a equals 116.52")[$a = 116.52$], #math.equation(block: false, alt: "A equals 85.07 degrees")[$" " A = 85.07^(∘)$], #math.equation(block: false, alt: "C equals 56.93 degrees")[$" " C = 56.93^(∘)$] or #math.equation(block: false, alt: "a equals 37.93")[$a = 37.93$], #math.equation(block: false, alt: "A equals 18.93 degrees")[$" " A = 18.93^(∘)$], #math.equation(block: false, alt: "C equals 123.07 degrees")[$" " C = 123.07^(∘)$] #math.equation(block: true, alt: "a equals 28 , c equals 41 , A equals 27 degrees")[$a = 28 , " " c = 41 , " " A = 27^(∘)$] #math.equation(block: true, alt: "c equals 5.7 , A equals 59 degrees , B equals 82 degrees")[$c = 5.7 , " " A = 59^(∘) , " " B = 82^(∘)$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "a equals 7.76")[$a = 7.76$], #math.equation(block: false, alt: "b equals 8.97")[$" " b = 8.97$], #math.equation(block: false, alt: "C equals 39 degrees")[$" " C = 39^(∘)$] #math.equation(block: true, alt: "b equals 82 , A equals 11 degrees , C equals 42 degrees")[$b = 82 , " " A = 11^(∘) , " " C = 42^(∘)$] For Problems 43–52, + Sketch and label a triangle to illustrate the problem. + Solve the problem. Round answers to one decimal place. A surveyor would like to know the distance #math.equation(block: false, alt: "P Q")[$P Q$] across a small lake, as shown in the figure. She stands at point #math.equation(block: false, alt: "O")[$O$] and measures the angle between the lines of sight to points #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$] at #math.equation(block: false, alt: "76 degrees")[$76^(∘)$]. She also finds #math.equation(block: false, alt: "O P equals 1400")[$O P = 1400$] meters and #math.equation(block: false, alt: "O Q equals 600")[$O Q = 600$] meters. Calculate the distance #math.equation(block: false, alt: "P Q")[$P Q$]. #figure(figph[lake], alt: "lake", caption: none) + #figure(figph[triangle], alt: "triangle", caption: none) + 1383.3 m Highway engineers plan to drill a tunnel through Boney Mountain from #math.equation(block: false, alt: "G")[$G$] to #math.equation(block: false, alt: "H")[$H$], as shown in the figure. The angle at point #math.equation(block: false, alt: "F")[$F$] is #math.equation(block: false, alt: "41 degrees")[$41^(∘)$], and the distances to #math.equation(block: false, alt: "G")[$G$] and #math.equation(block: false, alt: "H")[$H$] are 900 yards and 2500 yards, respectively. How long will the tunnel be? #figure(figph[mountains], alt: "mountains", caption: none) Two pilots leave an airport at the same time. One pilot flies #math.equation(block: false, alt: "3 degrees")[$3^(∘)$] east of north at a speed of 320 miles per hour, the other flies #math.equation(block: false, alt: "157 degrees")[$157^(∘)$] east of north at a speed of 406 miles per hour. How far apart are the two pilots after 3 hours? What is the heading from the first plane to the second plane at that time? + #figure(figph[triangle], alt: "triangle", caption: none) + 2123 mi, #math.equation(block: false, alt: "168.43 degrees")[$168.43^(∘)$] east of north Two boats leave port at the same time. One boat sails due west at a speed of 17 miles per hour, the other powers #math.equation(block: false, alt: "42 degrees")[$42^(∘)$] east of north at a speed of 23 miles per hour. How far apart are the two boats after 2 hours? What is the heading from the first boat to the second boat at that time? Caroline wants to fly directly south from Indianapolis to Cancun, Mexico, a distance of 1290 miles. However, to avoid bad weather, she flies for 400 miles on a heading #math.equation(block: false, alt: "18 degrees")[$18^(∘)$] east of south. What is the heading to Cancun from that location, and how far is it? + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "7.74 degrees")[$7.74^(∘)$] west of south, 917.9 mi Alex sails 8 miles from Key West, Florida on a heading #math.equation(block: false, alt: "40 degrees")[$40^(∘)$] east of south. He then changes course and sails for 10 miles due east. What is the heading back to Key West from that point, and how far is it? The phone company wants to erect a cell tower on a steep hill inclined #math.equation(block: false, alt: "26 degrees")[$26^(∘)$] to the horizontal. The installation crew plans to run a guy wire from a point on the ground 20 feet uphill from the base of the tower and attach it to the tower at a height of 100 feet. How long should the guy wire be? + #figure(figph[triangle], alt: "triangle", caption: none) + 92.99 ft Sandstone Peak rises 3500 feet above the desert. The Park Service plans to run an aerial tramway up the north face, which is inclined at an angle of #math.equation(block: false, alt: "68 degrees")[$68^(∘)$] to the horizontal. The base station will be located 500 feet from the foot of Sandstone Peak. Ignoring any slack in the cable, how long should it be? The sides of a triangle are 27 cm, 15 cm, and 20 cm. Find the area of the triangle. (Hint: Find one of the angles first.) #math.equation(block: true, alt: "147.73 cm squared")[$147.73 " " "cm"^(2)$] The sides of a parallelogram are 10 inches and 8 inches, and form an angle of #math.equation(block: false, alt: "130 degrees")[$130^(∘)$]. Find the lengths of the diagonals of the parallelogram. For Problems 53–56, find #math.equation(block: false, alt: "x")[$x$], the distance from one vertex to the foot of the altitude. #figure(figph[triangle], alt: "triangle", caption: none) 10.53 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) 4.08 #figure(figph[triangle], alt: "triangle", caption: none) Problems 57 and 58 prove the Law of Cosines. #figure(figph[triangles], alt: "triangles", caption: none) + Copy the three figures above showing the three possibilities for an angle #math.equation(block: false, alt: "C")[$C$] in a triangle: #math.equation(block: false, alt: "C")[$C$] is acute, obtuse, or a right angle. For each figure, explain why it is true that #math.equation(block: false, alt: "c squared equals open parenthesis b minus x close parenthesis squared plus y squared")[$c^(2) = ( b − x )^(2) + y^(2)$], then rewrite the right side to get #math.equation(block: false, alt: "c squared equals open parenthesis x squared plus y squared close parenthesis plus b squared minus 2 b x")[$c^(2) = ( x^(2) + y^(2) ) + b^(2) − 2 b x$]. + For each figure, explain why it is true that #math.equation(block: false, alt: "x squared plus y squared equals a squared")[$x^(2) + y^(2) = a^(2)$]. + For all three figures, #math.equation(block: false, alt: "a")[$a$] is the distance from the origin to the point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$]. Use the definition of cosine to write #math.equation(block: false, alt: "cos ⁡ open parenthesis C close parenthesis")[$cos ( C )$] in terms of #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "x")[$x$], then solve your equation for #math.equation(block: false, alt: "x")[$x$]. + Start with the last equation from (a), and substitute expressions from (b) and (c) to conclude one case of the Law of Cosines. + First figure: #math.equation(block: false, alt: "b minus x")[$b − x$] is the base of the small right triangle. Second: #math.equation(block: false, alt: "minus x")[$− x$] is the horizontal distance between #math.equation(block: false, alt: "P")[$P$] and the #math.equation(block: false, alt: "x")[$x$]-axis, so #math.equation(block: false, alt: "b plus open parenthesis minus x close parenthesis")[$b + ( − x )$] or #math.equation(block: false, alt: "b minus x")[$b − x$] is the base of the large right triangle. Third: #math.equation(block: false, alt: "x equals 0")[$x = 0$], and #math.equation(block: false, alt: "b")[$b$] is the base of a right triangle. + First: #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] are the legs of a right triangle, #math.equation(block: false, alt: "a")[$a$] is the hypotenuse. Second: #math.equation(block: false, alt: "minus x")[$− x$] and #math.equation(block: false, alt: "y")[$y$] are the legs of a right triangle with hypotenuse #math.equation(block: false, alt: "a")[$a$]. Third: #math.equation(block: false, alt: "x equals 0")[$x = 0$] and #math.equation(block: false, alt: "y equals a")[$y = a$] + #math.equation(block: false, alt: "x equals a cos ⁡ open parenthesis C close parenthesis")[$x = a cos ( C )$] Demonstrate the other two cases of the Law of Cosines: - #math.equation(block: false, alt: "a squared equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis")[$a^(2) = b^(2) + c^(2) − 2 b c cos ( A )$] - #math.equation(block: false, alt: "b squared equals a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis")[$b^(2) = a^(2) + c^(2) − 2 a c cos ( B )$] (Hint: See Problem 57 and switch the roles of #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "c")[$c$], etc.) Use the Law of Cosines to prove the projection laws: #math.equation(block: true, alt: "a, equals b cos ⁡ open parenthesis C close parenthesis plus c cos ⁡ open parenthesis B close parenthesis; b, equals c cos ⁡ open parenthesis A close parenthesis plus a cos ⁡ open parenthesis C close parenthesis; c, equals a cos ⁡ open parenthesis B close parenthesis plus b cos ⁡ open parenthesis A close parenthesis")[$a & = b cos ( C ) + c cos ( B ) \ b & = c cos ( A ) + a cos ( C ) \ c & = a cos ( B ) + b cos ( A )$] Illustrate with a sketch. (Hint: Add together two of the versions of the Law of Cosines.) #math.equation(block: true, alt: "b squared plus c squared, equals open parenthesis a squared plus c squared minus 2 a c cos ⁡ open parenthesis B close parenthesis close parenthesis plus open parenthesis a squared plus b squared minus 2 b c cos ⁡ open parenthesis C close parenthesis close parenthesis; equals 2 a squared plus b squared plus c squared minus 2 a open parenthesis c cos ⁡ open parenthesis B close parenthesis plus b cos ⁡ open parenthesis C close parenthesis close parenthesis")[$b^(2) + c^(2) & = ( a^(2) + c^(2) − 2 a c cos ( B ) ) + ( a^(2) + b^(2) − 2 b c cos ( C ) ) \ & = 2 a^(2) + b^(2) + c^(2) − 2 a ( c cos ( B ) + b cos ( C ) )$] so #math.equation(block: false, alt: "2 a squared equals 2 a open parenthesis c cos ⁡ open parenthesis B close parenthesis plus b cos ⁡ open parenthesis C close parenthesis close parenthesis")[$2 a^(2) = 2 a ( c cos ( B ) + b cos ( C ) )$], and dividing both sides by #math.equation(block: false, alt: "2 a")[$2 a$] yields #math.equation(block: false, alt: "a equals open parenthesis c cos ⁡ open parenthesis B close parenthesis plus b cos ⁡ open parenthesis C close parenthesis")[$a = \( c cos \( B \) + b cos \( C \)$] If #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] is isosceles with #math.equation(block: false, alt: "a equals b")[$a = b$], show that #math.equation(block: false, alt: "c squared equals 2 a squared open parenthesis 1 minus cos ⁡ open parenthesis C close parenthesis close parenthesis")[$c^(2) = 2 a^(2) ( 1 − cos ( C ) )$]. Use the Law of Cosines to prove: #math.equation(block: true, alt: "1 plus cos ⁡ open parenthesis A close parenthesis, equals the fraction open parenthesis a plus b plus c close parenthesis open parenthesis minus a plus b plus C close parenthesis over 2 b c; 1 minus cos ⁡ open parenthesis A close parenthesis, equals the fraction open parenthesis a minus b plus c close parenthesis open parenthesis minus a plus b plus C close parenthesis over 2 b c")[$1 + cos ( A ) & = display(frac(( a + b + c ) ( − a + b + C ), 2 b c)) \ 1 − cos ( A ) & = display(frac(( a − b + c ) ( − a + b + C ), 2 b c))$] For the first equation, start with the Law of Cosines in the form #math.equation(block: true, alt: "a squared equals b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis")[$a^(2) = b^(2) + c^(2) − 2 b c cos ( A )$] Add #math.equation(block: false, alt: "2 a b plus 2 b c cos ⁡ open parenthesis A close parenthesis minus a squared")[$2 a b + 2 b c cos ( A ) − a^(2)$] to both sides of the equation, factor the right side, then divide both sides by #math.equation(block: false, alt: "2 b c")[$2 b c$]. For the second equation, start with the Law of Cosines in the form #math.equation(block: true, alt: "b squared plus c squared minus 2 b c cos ⁡ open parenthesis A close parenthesis equals a squared")[$b^(2) + c^(2) − 2 b c cos ( A ) = a^(2)$] Add #math.equation(block: false, alt: "2 b c minus b squared minus c squared")[$2 b c − b^(2) − c^(2)$] to both sides of the equation, factor the right side, then divide both sides by #math.equation(block: false, alt: "2 b c")[$2 b c$]. Prove that #math.equation(block: true, alt: "the fraction cos ⁡ open parenthesis A close parenthesis over a plus the fraction cos ⁡ open parenthesis B close parenthesis over b plus the fraction cos ⁡ open parenthesis C close parenthesis over c equals the fraction a squared plus b squared plus c squared over 2 a b c")[$display(frac(cos ( A ), a)) + display(frac(cos ( B ), b)) + display(frac(cos ( C ), c)) = display(frac(a^(2) + b^(2) + c^(2), 2 a b c))$]