#set document(title: "3.2 The Law of Sines", 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")) == 3.2#h(0.6em)The Law of Sines We have learned to use the trigonometric ratios to solve right triangles. But the trig ratios are only valid for the sides of right triangles. Can we find unknown sides or angles in an oblique triangle? #figure(figph[rigth triangle, oblique triangle], alt: "rigth triangle, oblique triangle", caption: none) In this section and the next we find relationships among the sides and angles of oblique triangles. These relationships are called the Law of Sines and the Law of Cosines. To derive these new rules, we use what we already know about right triangles. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Reducing a new problem to an earlier one is a frequently used technique in mathematics. ] Consider the oblique triangle below. By drawing in the altitude #math.equation(block: false, alt: "h")[$h$] of the triangle, we create two right triangles, #math.equation(block: false, alt: "△ B C D")[$upright(△) B C D$] and #math.equation(block: false, alt: "△ A B D")[$upright(△) A B D$], as shown in the figure. Now we can write expressions in terms of #math.equation(block: false, alt: "h")[$h$] for #math.equation(block: false, alt: "sin ⁡ A")[$sin A$] and for #math.equation(block: false, alt: "sin ⁡ C")[$sin C$]. Looking at #math.equation(block: false, alt: "△ B C D")[$upright(△) B C D$], we see that #math.equation(block: true, alt: "the fraction h over a equals sin ⁡ open parenthesis C close parenthesis .")[$display(frac(h, a)) = sin ( C ) .$] Looking at #math.equation(block: false, alt: "△ A B D")[$upright(△) A B D$], we see that #math.equation(block: true, alt: "the fraction h over c equals sin ⁡ open parenthesis A close parenthesis .")[$display(frac(h, c)) = sin ( A ) .$] #figure(figph[oblique triangle with altitude], alt: "oblique triangle with altitude", caption: none) Now we solve each of these equations for #math.equation(block: false, alt: "h")[$h$]: #math.equation(block: true, alt: "the fraction h over a, equals sin ⁡ open parenthesis C close parenthesis, and, the fraction h over c, equals sin ⁡ open parenthesis A close parenthesis, Solve each equation for h .; h, equals a sin ⁡ open parenthesis C close parenthesis, and, h, equals c sin ⁡ open parenthesis A close parenthesis, Equate the expressions for h .")[$display(frac(h, a)) & = sin ( C ) & & "and" & display(frac(h, c)) & = sin ( A ) & & "Solve each equation for " bold(italic(h)) . \ h & = a sin ( C ) & & "and" & h & = c sin ( A ) & & "Equate the expressions for " bold(italic(h)) .$] #math.equation(block: true, alt: "a sin ⁡ open parenthesis C close parenthesis, equals c sin ⁡ open parenthesis A close parenthesis, Divide both sides by a c .; the fraction sin ⁡ open parenthesis C close parenthesis over c, equals the fraction sin ⁡ open parenthesis A close parenthesis over a")[$& & a sin ( C ) & = c sin ( A ) & & #h(2em) #h(1em) & & "Divide both sides by " bold(italic(a)) bold(italic(c)) . \ & & display(frac(sin ( C ), c)) & = display(frac(sin ( A ), a))$] We have derived a relationship between the angles #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "C")[$C$] and their opposite sides, #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "c")[$c$]. If we know any three of these quantities, we can find the fourth. In a similar way, by drawing in the altitude from the vertex #math.equation(block: false, alt: "C")[$C$], we can show that #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis A close parenthesis over a equals the fraction sin ⁡ open parenthesis B close parenthesis over b")[$display(frac(sin ( A ), a)) = display(frac(sin ( B ), b))$] Putting both results together, we have the #strong[Law of Sines]. The Law of Sines is true for any triangle, whether it is acute, right, or obtuse. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Law of Sines] 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: "the fraction sin ⁡ open parenthesis A close parenthesis over a equals the fraction sin ⁡ open parenthesis B close parenthesis over b equals the fraction sin ⁡ open parenthesis C close parenthesis over c")[$display(frac(bold(sin) ( bold(italic(A)) ), bold(italic(a)))) = display(frac(bold(sin) ( bold(italic(B)) ), bold(italic(b)))) = display(frac(bold(sin) ( bold(italic(C)) ), bold(italic(c))))$] #figure(figph[triangle ABC], alt: "triangle ABC", caption: none) or equivalently, #math.equation(block: true, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis equals the fraction b over sin ⁡ open parenthesis B close parenthesis equals the fraction c over sin ⁡ open parenthesis C close parenthesis")[$display(frac(bold(italic(a)), bold(sin) ( bold(italic(A)) ))) = display(frac(bold(italic(b)), bold(sin) ( bold(italic(B)) ))) = display(frac(bold(italic(c)), bold(sin) ( bold(italic(C)) )))$] ] === Finding a Side In the next example, we use the Law of Sines to find a distance. #examplebox("Example 1")[][ Two observers onshore sight a ship at an unknown distance from the shore. The observers are 400 yards apart at points #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], and they each measure the angle from the shoreline to the ship, as shown below. How far is the ship from the observer at #math.equation(block: false, alt: "A")[$A$]? #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ First note that #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] is not a right triangle, so we cannot use the trig ratios directly to find the sides of this triangle. The unknown distance #math.equation(block: false, alt: "d")[$d$] is the side opposite #math.equation(block: false, alt: "∠ B equals 79.4 degrees")[$upright(∠) B = 79.4^(∘)$]. In order to use the Law of Sines, we must know another angle and the side opposite that angle. We do: we know that side #math.equation(block: false, alt: "c equals 400")[$c = 400$], and we can compute the angle at the ship, #math.equation(block: false, alt: "∠ C")[$upright(∠) C$]. #math.equation(block: true, alt: "∠ C equals 180 degrees minus open parenthesis 79.4 degrees plus 83.2 degrees close parenthesis equals 17.4 degrees")[$upright(∠) C = 180^(∘) − ( 79.4^(∘) + 83.2^(∘) ) = 17.4^(∘)$] Now we apply the Law of Sines, using angles #math.equation(block: false, alt: "B")[$B$] and #math.equation(block: false, alt: "C")[$C$]. #math.equation(block: true, alt: "the fraction b over sin ⁡ open parenthesis B close parenthesis, equals the fraction c over sin ⁡ open parenthesis C close parenthesis, Substitute the given values.; the fraction d over sin ⁡ open parenthesis 79.4 degrees close parenthesis, equals the fraction 400 over sin ⁡ open parenthesis 17.4 degrees close parenthesis, Evaluate sines.; the fraction d over 0.9829, equals the fraction 400 over 0.2990, Multiply both sides by 0.9829.; d, equals 1315")[$display(frac(b, sin ( B ))) & = display(frac(c, sin ( C ))) & & "Substitute the given values." \ display(frac(d, sin ( 79.4^(∘) ))) & = display(frac(400, sin ( 17.4^(∘) ))) & & "Evaluate sines." \ display(frac(d, 0.9829)) & = display(frac(400, 0.2990)) & & "Multiply both sides by 0.9829." \ d & = 1315$] The ship is about 1315 yards from the observer at #math.equation(block: false, alt: "A")[$A$]. ] ] Delbert and Francine are 40 feet apart on one side of a river. They make angle measurements to a pine tree on the opposite shore as shown below. What is the distance from Francine to the pine tree? #figure(figph[triangle], alt: "triangle", caption: none) The angle at #math.equation(block: false, alt: "P")[$P$] is #math.equation(block: false, alt: "180 degrees minus open parenthesis 79 degrees plus 81 degrees close parenthesis equals 20 degrees")[$180^(∘) − ( 79^(∘) + 81^(∘) ) = 20^(∘)$]. Now we can use the Law of Sines: #math.equation(block: true, alt: "the fraction d over sin ⁡ open parenthesis 79 degrees close parenthesis equals the fraction 40 over sin ⁡ open parenthesis 20 degrees close parenthesis")[$display(frac(d, sin ( 79^(∘) ))) = display(frac(40, sin ( 20^(∘) )))$] Solving for #math.equation(block: false, alt: "d")[$d$] we find the distance is about 114.8 feet. === Solving Triangles with the Law of Sines In order to apply the Law of Sines to find a side, we must know one angle of the triangle and its opposite side (either #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "A")[$A$], or #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "B")[$B$], or #math.equation(block: false, alt: "c")[$c$] and #math.equation(block: false, alt: "C")[$C$]), and one other angle. Then we can find the side opposite that angle. #examplebox("Example 2")[][ In the triangle shown at right, #math.equation(block: true, alt: "A equals 37 degrees , B equals 54 degrees , and a equals 11")[$A = 37^(∘) , B = 54^(∘) , " " "and" " " a = 11$] + Find #math.equation(block: false, alt: "b")[$b$]. + Solve the triangle. #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ + We use the Law of Sines with #math.equation(block: false, alt: "a")[$a$] and angle #math.equation(block: false, alt: "A")[$A$] to find #math.equation(block: false, alt: "b")[$b$]. #math.equation(block: true, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis, equals the fraction b over sin ⁡ open parenthesis B close parenthesis, Substitute the given values.; the fraction 11 over sin ⁡ open parenthesis 37 degrees close parenthesis, equals the fraction b over sin ⁡ open parenthesis 54 degrees close parenthesis, Solve for b .; the fraction 11 over sin ⁡ open parenthesis 37 degrees close parenthesis times sin ⁡ open parenthesis 54 degrees close parenthesis, equals b")[$display(frac(a, sin ( A ))) & = display(frac(b, sin ( B ))) & & "Substitute the given values." \ display(frac(11, sin ( 37^(∘) ))) & = display(frac(b, sin ( 54^(∘) ))) & & "Solve for " bold(italic(b)) . \ display(frac(11, sin ( 37^(∘) ))) ⋅ sin ( 54^(∘) ) & = b$] Evaluating this expression with a calculator, we find that #math.equation(block: false, alt: "b approximately equals 14.79")[$b ≈ 14.79$]. + The angle #math.equation(block: false, alt: "C equals 180 degrees minus open parenthesis 37 degrees plus 54 degrees close parenthesis equals 89 degrees")[$" " C = 180^(∘) − ( 37^(∘) + 54^(∘) ) = 89^(∘)$]. Now we use the Law of Sines to find side #math.equation(block: false, alt: "c")[$c$]. Note that it is safer to use the given side, #math.equation(block: false, alt: "a")[$a$], rather than the value we calculated for #math.equation(block: false, alt: "b")[$b$]. #math.equation(block: true, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis, equals the fraction c over sin ⁡ open parenthesis C close parenthesis, Substitute the known values.; the fraction 11 over sin ⁡ open parenthesis 37 degrees close parenthesis, equals the fraction c over sin ⁡ open parenthesis 89 degrees close parenthesis, Solve for c .; the fraction 11 over sin ⁡ open parenthesis 37 degrees close parenthesis times sin ⁡ open parenthesis 89 degrees close parenthesis, equals c")[$display(frac(a, sin ( A ))) & = display(frac(c, sin ( C ))) & & "Substitute the known values." \ display(frac(11, sin ( 37^(∘) ))) & = display(frac(c, sin ( 89^(∘) ))) & & "Solve for " bold(italic(c)) . \ display(frac(11, sin ( 37^(∘) ))) ⋅ sin ( 89^(∘) ) & = c$] so #math.equation(block: false, alt: "c approximately equals 18.28.")[$c ≈ 18.28 .$] ] ] In the triangle at right, #math.equation(block: true, alt: "A equals 65 degrees , C equals 42 degrees , and c equals 16")[$A = 65^(∘) , C = 42^(∘) , " " "and" " " c = 16$] Solve the triangle. (Hint: Which side will you find first?) #figure(figph[triangle], alt: "triangle", caption: none) First use the Law of Sines to find #math.equation(block: false, alt: "a")[$a$]. #math.equation(block: true, alt: "the fraction a over sin ⁡ open parenthesis 65 degrees close parenthesis equals the fraction 16 over sin ⁡ open parenthesis 42 degrees close parenthesis")[$display(frac(a, sin ( 65^(∘) ))) = display(frac(16, sin ( 42^(∘) )))$] Solving for #math.equation(block: false, alt: "a")[$a$] we find #math.equation(block: false, alt: "a equals 21.67")[$a = 21.67$]. Next we find angle #math.equation(block: false, alt: "B")[$B$]. #math.equation(block: true, alt: "b equals 180 degrees minus open parenthesis 65 degrees plus 42 degrees close parenthesis equals 73 degrees")[$b = 180^(∘) − ( 65^(∘) + 42^(∘) ) = 73^(∘)$] Finally, we use the Law of Sines to find #math.equation(block: false, alt: "b")[$b$]. #math.equation(block: true, alt: "the fraction b over sin ⁡ open parenthesis 73 degrees close parenthesis equals the fraction 16 over sin ⁡ open parenthesis 42 degrees close parenthesis")[$display(frac(b, sin ( 73^(∘) ))) = display(frac(16, sin ( 42^(∘) )))$] Solving this equation, we find #math.equation(block: false, alt: "b equals 22.87")[$b = 22.87$]. === Finding an Angle We can also use the Law of Sines to find an unknown angle of a triangle. We must know two sides of the triangle and the angle opposite one of them. #examplebox("Example 3")[][ In the triangle shown at right, #math.equation(block: true, alt: "B equals 55 degrees , a equals 5 , and b equals 11")[$B = 55^(∘) , a = 5 , " " "and" " " b = 11$] Solve the triangle. #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We must find the three remaining parts of the triangle, #math.equation(block: false, alt: "A , C")[$A , C$], and #math.equation(block: false, alt: "c")[$c$]. First, we use the Law of Sines to find #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis")[$sin ( A )$]. #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis A close parenthesis over a, equals the fraction sin ⁡ open parenthesis B close parenthesis over b, Substitute the known values.; the fraction sin ⁡ open parenthesis A close parenthesis over 5, equals the fraction sin ⁡ open parenthesis 55 degrees close parenthesis over 11, Multiply both sides by 5.; sin ⁡ open parenthesis A close parenthesis, equals 5 times the fraction sin ⁡ open parenthesis 55 degrees close parenthesis over 11 approximately equals 0.3723")[$display(frac(sin ( A ), a)) & = display(frac(sin ( B ), b)) & & "Substitute the known values." \ display(frac(sin ( A ), 5)) & = display(frac(sin ( 55^(∘) ), 11)) & & "Multiply both sides by 5." \ sin ( A ) & = 5 ⋅ display(frac(sin ( 55^(∘) ), 11)) ≈ 0.3723$] So #math.equation(block: false, alt: "A equals sin to the power minus 1 ⁡ open parenthesis 0.3723 close parenthesis approximately equals 21.9 degrees")[$" " A = sin^(− 1) ( 0.3723 ) ≈ 21.9^(∘)$]. Now we know two angles, and we can find angle #math.equation(block: false, alt: "C")[$C$]. #math.equation(block: true, alt: "C, equals 180 degrees minus open parenthesis B plus A close parenthesis; equals 180 degrees minus 55 degrees minus 21.9 degrees equals 103.1 degrees")[$C & = 180^(∘) − ( B + A ) \ & = 180^(∘) − 55^(∘) − 21.9^(∘) = 103.1^(∘)$] Finally, we use the Law of Sines to find side #math.equation(block: false, alt: "c")[$c$]. #math.equation(block: true, alt: "the fraction c over sin ⁡ open parenthesis C close parenthesis, equals the fraction b over sin ⁡ open parenthesis B close parenthesis, Substitute the known values.; the fraction c over sin ⁡ open parenthesis 103.1 degrees close parenthesis, equals the fraction 11 over sin ⁡ open parenthesis 55 degrees close parenthesis, Multiply both sides by; sin ⁡ open parenthesis 103.1 degrees close parenthesis .; c, equals the fraction 11 over sin ⁡ open parenthesis 55 degrees close parenthesis times sin ⁡ open parenthesis 103.1 degrees close parenthesis")[$display(frac(c, sin ( C ))) & = display(frac(b, sin ( B ))) & & "Substitute the known values." \ display(frac(c, sin ( 103.1^(∘) ))) & = display(frac(11, sin ( 55^(∘) ))) & & "Multiply both sides by " \ bold(sin) ( 103.1^(∘) ) . \ " " c & = display(frac(11, sin ( 55^(∘) ))) ⋅ sin ( 103.1^(∘) )$] Evaluating this expression with a calculator gives #math.equation(block: false, alt: "c approximately equals 13.1.")[$c ≈ 13.1 .$] ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the last Example we used the value of #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis")[$sin ( A )$] to find angle #math.equation(block: false, alt: "A")[$A$]. But there are two angles that satisfy #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals 0.3723")[$sin ( A ) = 0.3723$], one acute (21.9#math.equation(block: false, alt: "∘")[$∘$]) and one obtuse (158.1#math.equation(block: false, alt: "∘")[$∘$]). How did we know which one to choose? #figure(figph[supplementary angles], alt: "supplementary angles", caption: none) In this case, the obtuse angle is too big to fit in the triangle, because angle #math.equation(block: false, alt: "B equals 55 degrees")[$B = 55^(∘)$]. Sometimes both angles will produce (different) triangles, and sometimes only the acute angle will work. You should always check whether both angles provide solutions. ] Sketch a triangle with #math.equation(block: false, alt: "C equals 93 degrees , a equals 7")[$C = 93^(∘) , a = 7$], and #math.equation(block: false, alt: "c equals 11")[$c = 11$]. + Use the Law of Sines to find another angle of the triangle. + Solve the triangle, and label your sketch with the results. + We can use the Law of Sines to find angle #math.equation(block: false, alt: "A")[$A$]. #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis A close parenthesis over 7 equals the fraction sin ⁡ open parenthesis 93 degrees close parenthesis over 11")[$display(frac(sin ( A ), 7)) = display(frac(sin ( 93^(∘) ), 11))$] Solve for #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis")[$sin ( A )$] to find #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis equals 0.6355")[$sin ( A ) = 0.6355$], so #math.equation(block: true, alt: "A equals sin to the power minus 1 ⁡ open parenthesis 0.6355 close parenthesis equals 39.5 degrees")[$A = sin^(− 1) ( 0.6355 ) = 39.5^(∘)$]. (Note that the supplement of #math.equation(block: false, alt: "39.5 degrees")[$39.5^(∘)$], or #math.equation(block: false, alt: "140.5 degrees")[$140.5^(∘)$], is too large to fit in the triangle because #math.equation(block: false, alt: "C equals 93 degrees")[$C = 93^(∘)$].) + Now we can find angle #math.equation(block: false, alt: "B equals 47.5 degrees")[$B = 47.5^(∘)$], and use the Law of Sines again to find #math.equation(block: false, alt: "b equals 8.13")[$b = 8.13$]. #examplebox("Example 4")[][ Find two triangles in which #math.equation(block: false, alt: "B equals 14.4 degrees , a equals 8")[$B = 14.4^(∘) , a = 8$], and #math.equation(block: false, alt: "b equals 3")[$b = 3$], and sketch both triangles. #solutionbox[ Using the Law of Sines, we have #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis A close parenthesis over a, equals the fraction sin ⁡ open parenthesis B close parenthesis over b, Substitute the known values.; the fraction sin ⁡ open parenthesis A close parenthesis over 8, equals the fraction sin ⁡ open parenthesis 14.4 degrees close parenthesis over 3, Multiply both sides by 8.; sin ⁡ open parenthesis A close parenthesis, equals 8 times the fraction sin ⁡ open parenthesis 14.4 degrees close parenthesis over 3 approximately equals 0.6632")[$display(frac(sin ( A ), a)) & = display(frac(sin ( B ), b)) & & "Substitute the known values." \ display(frac(sin ( A ), 8)) & = display(frac(sin ( 14.4^(∘) ), 3)) & & "Multiply both sides by 8." \ sin ( A ) & = 8 ⋅ display(frac(sin ( 14.4^(∘) ), 3)) ≈ 0.6632$] There are two angles with sine #math.equation(block: false, alt: "0.6632")[$0.6632$]: - the acute angle #math.equation(block: false, alt: "A equals sin to the power minus 1 ⁡ open parenthesis 0.6632 close parenthesis equals 41.5 degrees ,")[$" " A = sin^(− 1) ( 0.6632 ) = 41.5^(∘) ,$] - or its supplement, #math.equation(block: false, alt: "A prime equals 180 degrees minus 41.5 degrees equals 138.5 degrees")[$" " A^(′) = 180^(∘) − 41.5^(∘) = 138.5^(∘)$]. Each of these angles produces a different solution triangle, because the angle #math.equation(block: false, alt: "C")[$C$] and side #math.equation(block: false, alt: "c")[$c$] will be different also. - In the first case, angle #math.equation(block: false, alt: "C equals 180 degrees minus open parenthesis 14.4 degrees plus 138.5 degrees close parenthesis equals 27.1 degrees")[$" " C = 180^(∘) − ( 14.4^(∘) + 138.5^(∘) ) = 27.1^(∘)$], and we have the triangle shown in figure (a). - In the second case, #math.equation(block: false, alt: "C equals 180 degrees minus open parenthesis 14.4 degrees plus 41.5 degrees close parenthesis equals 124.1 degrees")[$" " C = 180^(∘) − ( 14.4^(∘) + 41.5^(∘) ) = 124.1^(∘)$], which gives us the triangle shown in figure (b). #figure(figph[triangles], alt: "triangles", caption: none) ] ] Suppose that #math.equation(block: false, alt: "C equals 29.7 degrees , b equals 8")[$" " C = 29.7^(∘) , " " b = 8$], and #math.equation(block: false, alt: "c equals 5")[$" " c = 5$]. + Find two possible values for angle #math.equation(block: false, alt: "B")[$B$]. + Solve the triangle for both values of #math.equation(block: false, alt: "B")[$B$], and sketch both solutions. + From the Law of Sines we find #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis B close parenthesis over 8 equals the fraction sin ⁡ open parenthesis 29.7 degrees close parenthesis over 5")[$display(frac(sin ( B ), 8)) = display(frac(sin ( 29.7^(∘) ), 5))$] so #math.equation(block: false, alt: "sin ⁡ B equals 0.7927")[$" " sin B = 0.7927 " "$] and #math.equation(block: false, alt: "B equals 52.4 degrees")[$B = 52.4^(∘)$] or #math.equation(block: false, alt: "B equals 127.6 degrees")[$B = 127.6^(∘)$] + In each case, we find angle #math.equation(block: false, alt: "A")[$A$] and then use the Law of Sines to find side #math.equation(block: false, alt: "a")[$a$]. #math.equation(block: true, alt: "A equals 97.9 degrees , a equals 10")[$A = 97.9^(∘) , " " a = 10$] #figure(figph[triangle], alt: "triangle", caption: none) or #math.equation(block: false, alt: "A equals 22.7 degrees , a equals 3.9")[$A = 22.7^(∘) , " " a = 3.9$] #figure(figph[triangle], alt: "triangle", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Because there can be more than one solution for a triangle in which we know two sides and an angle opposite one of them, this situation is called the #strong[ambiguous case] for the Law of Sines. (See Homework Problems 33–42 for more about the ambiguous case.) Side b = 5 is fixed, running from vertex A at the chosen angle up to vertex C; the dashed segment is the height h = 5 sin A. The red circle has radius a, the side opposite A, and vertex B must lie where that circle meets the baseline — so counting the crossings counts the triangles. Set A = 30°, which makes h exactly 2.5, and sweep a through all four outcomes: at a = 1.5 the circle floats clear of the baseline and NO triangle exists; at a = 2.5 it just touches, giving exactly one right triangle; anywhere between 3 and 4.5 it cuts the baseline twice, so two genuinely different triangles share the same A, b and a — the ambiguous case this sentence names; and from a = 5 upward it crosses once on the far side only, leaving one triangle again. The angle stops at 80° because the ambiguity is a property of acute A: once A is a right angle the height equals b and the tangent case collapses to a single point rather than a triangle.{"functions":\[{"color":"\#6a6a6a","expression":"12\*t","expression2":"0","id":"baseline","inequality":"lt","lineDash":\[\],"lineWidth":1.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*5\*cos(A\*pi/180)","expression2":"t\*5\*sin(A\*pi/180)","id":"side-b","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"5\*cos(A\*pi/180)","expression2":"t\*5\*sin(A\*pi/180)","id":"height","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"5\*cos(A\*pi/180) + a\*cos(t)","expression2":"5\*sin(A\*pi/180) + a\*sin(t)","id":"swing-circle","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"5\*cos(A\*pi/180) + 0.13\*cos(t)","expression2":"5\*sin(A\*pi/180) + 0.13\*sin(t)","id":"vertex-c","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-A","label":"Angle A at the origin","max":80,"min":20,"name":"A","step":5,"unit":"\\u00b0","value":30},{"id":"p-a","label":"Side a opposite A \\u2014 the circle's radius","max":6,"min":1,"name":"a","step":0.5,"value":3}\],"title":"SSA: how many triangles have this A, b and a?","version":1,"viewport":{"centerX":4,"centerY":1.5,"scale":46}} ] === Applications In the next example, we use two triangles to solve the problem. #examplebox("Example 5")[][ Richard wants to measure the height of a castle controlled by hostile forces. When he is as close as he can get to the castle, the angle of elevation to the top of the wall is #math.equation(block: false, alt: "18.5 degrees")[$18.5^(∘)$]. He then retreats 20 yards and measures the angle of elevation again; this time it is #math.equation(block: false, alt: "15.9 degrees")[$15.9^(∘)$]. How tall is the castle? #figure(figph[castle with angles of elevation], alt: "castle with angles of elevation", caption: none) #solutionbox[ Notice that #math.equation(block: false, alt: "h")[$h$] is one side of the right triangle #math.equation(block: false, alt: "A D C")[$A D C$]. If we can find its hypotenuse, labeled #math.equation(block: false, alt: "r")[$r$] in the figure, we can use the sine ratio to find #math.equation(block: false, alt: "h")[$h$]. To find #math.equation(block: false, alt: "r")[$r$], we consider a second triangle, #math.equation(block: false, alt: "A B C")[$A B C$], as shown below. #figure(figph[triangles], alt: "triangles", caption: none) In this triangle, we know side #math.equation(block: false, alt: "B C equals 20")[$B C = 20$] and would like to find side #math.equation(block: false, alt: "A C equals r")[$A C = r$]. We can use the Law of Sines to find #math.equation(block: false, alt: "r")[$r$], but first we must calculate the other angles of the triangle. Now, the angle opposite #math.equation(block: false, alt: "r , ∠ A B C")[$r , " " upright(∠) A B C$], is the complement of #math.equation(block: false, alt: "18.5 degrees")[$18.5^(∘)$], so #math.equation(block: true, alt: "∠ A B C equals 180 degrees minus 18.5 degrees equals 161.5 degrees")[$upright(∠) A B C = 180^(∘) − 18.5^(∘) = 161.5^(∘)$] The angle opposite the 20-yard side, #math.equation(block: false, alt: "∠ B A C")[$upright(∠) B A C$], is #math.equation(block: true, alt: "∠ B A C equals 180 degrees minus open parenthesis 161.5 degrees plus 15.9 degrees close parenthesis equals 2.6 degrees")[$upright(∠) B A C = 180^(∘) − ( 161.5^(∘) + 15.9^(∘) ) = 2.6^(∘)$] Now we can apply the Law of Sines to find #math.equation(block: false, alt: "r")[$r$]. We have #math.equation(block: true, alt: "the fraction r over sin ⁡ open parenthesis 161.5 degrees close parenthesis, equals the fraction 20 over sin ⁡ open parenthesis 2.6 degrees close parenthesis, Solve for r .; r, equals sin ⁡ open parenthesis 161.5 degrees close parenthesis times the fraction 20 over sin ⁡ 2.6 degrees; approximately equals 139.9")[$display(frac(r, sin ( 161.5^(∘) ))) & = display(frac(20, sin ( 2.6^(∘) ))) & & "Solve for " bold(italic(r)) . \ r & = sin ( 161.5^(∘) ) ⋅ display(frac(20, sin 2.6^(∘))) \ & ≈ 139.9$] So #math.equation(block: false, alt: "r")[$r$] is about 139.9 yards. Finally, using the right triangle #math.equation(block: false, alt: "A D C")[$A D C$] and the definition of sine, we can write #math.equation(block: true, alt: "the fraction h over r, equals sin ⁡ open parenthesis 15.9 close parenthesis degrees, Solve for h .; h, equals r times sin ⁡ open parenthesis 15.9 close parenthesis degrees approximately equals 38.33")[$display(frac(h, r)) & = sin ( 15.9 )^(∘) & & "Solve for " bold(italic(h)) . \ h & = r ⋅ sin ( 15.9 )^(∘) ≈ 38.33$] The castle is about 38.33 yards tall. ] ] Solve the problem in the previous example again, but instead of finding #math.equation(block: false, alt: "r")[$r$], find the length #math.equation(block: false, alt: "A B")[$A B$], and then use #math.equation(block: false, alt: "△ A B D")[$upright(△) A B D$] to find #math.equation(block: false, alt: "h")[$h$]. We know that angle #math.equation(block: false, alt: "A B C")[$A B C$] is the supplement of angle #math.equation(block: false, alt: "A B D")[$A B D$], or #math.equation(block: false, alt: "161.5 degrees")[$161.5^(∘)$]. We can use the Law of Sines to find side #math.equation(block: false, alt: "A B equals 120.79")[$A B = 120.79$], and then use the tangent of #math.equation(block: false, alt: "18.5 degrees")[$18.5^(∘)$] to find #math.equation(block: false, alt: "h")[$h$]. The castle is about 38.33 yards tall. === Measuring Astronomical Distances If you look at a nearby object and alternately close your left and right eyes, the object seems to jump in position. This apparent change occurs because your eyes are viewing the object from two different positions spaced several centimeters apart. If the object at point #math.equation(block: false, alt: "O")[$O$] is straight ahead of one eye, it appears to be at some angle #math.equation(block: false, alt: "p")[$p$] away from the line of sight of the other eye. The angle #math.equation(block: false, alt: "p")[$p$] is called the #strong[parallax] of the object. Use the figure below to see that #math.equation(block: false, alt: "p")[$p$] is also the angle between the directions to your two eyes when viewed from point #math.equation(block: false, alt: "O")[$O$]. (What fact from geometry justifies this statement?) #figure(figph[parallax], alt: "parallax", caption: none) Astronomers use parallax to determine the distance from earth to stars and other celestial objects. Two observers on Earth at a known distance apart both measure the direction to the star. The difference in angle between those two directions is the parallax. #examplebox("Example 6")[][ Astronomers 1000 kilometers apart observe an asteroid with a parallax of #math.equation(block: false, alt: "0.001 degrees")[$0.001^(∘)$]. How far is the asteroid from Earth? #solutionbox[ We let #math.equation(block: false, alt: "x")[$x$] represent the distance to the asteroid. The asteroid and the two observers make an isosceles triangle with base 1000 km and equal sides of length #math.equation(block: false, alt: "x")[$x$], as shown below. #figure(figph[parallax triangle], alt: "parallax triangle", caption: none) The base angles of the triangle are both #math.equation(block: false, alt: "the fraction 180 degrees minus 0.001 degrees over 2 equals 89.999 degrees")[$" " display(frac(180^(∘) − 0.001^(∘), 2)) = 89.999^(∘)$]. By the Law of Sines, #math.equation(block: true, alt: "the fraction x over sin ⁡ open parenthesis 89.999 degrees close parenthesis, equals the fraction 1000 over sin ⁡ open parenthesis 0.001 degrees close parenthesis, Solve for x .; x, equals sin ⁡ open parenthesis 89.999 degrees close parenthesis times the fraction 1000 over sin ⁡ open parenthesis 0.001 degrees close parenthesis; x, approximately equals 57 , 000 , 000")[$display(frac(x, sin ( 89.999^(∘) ))) & = display(frac(1000, sin ( 0.001^(∘) ))) & & "Solve for " bold(italic(x)) . \ x & = sin ( 89.999^(∘) ) ⋅ display(frac(1000, sin ( 0.001^(∘) ))) \ x & ≈ 57 , 000 , 000$] The asteroid is about 57 million kilometers from Earth (roughly one third of the distance to the Sun). ] ] Two observers 800 kilometers apart observe an object with a parallax of #math.equation(block: false, alt: "0.0005 degrees")[$0.0005^(∘)$]. How far is the object from Earth? The base angles of the parallax triangle are both #math.equation(block: false, alt: "the fraction 180 degrees minus 0.0005 degrees over 2 equals 89.99975 degrees")[$display(frac(180^(∘) − 0.0005^(∘), 2)) = 89.99975^(∘)$]. Now use the Law of Sines to find the equal sides of the triangle. The object is about 91,673,247 km from Earth. === Small Angles: Minutes and Seconds To obtain the most accurate parallax measurements, the distance between the two observers should be as large as possible. But even measured from opposite sides of Earth's orbit, stars outside the solar system have parallaxes much smaller than #math.equation(block: false, alt: "0.0001 degrees")[$0.0001^(∘)$]. In order to handle such small angles, we divide degrees into smaller units called #strong[minutes] and #strong[seconds]. One minute is #math.equation(block: false, alt: "the fraction 1 over 60")[$display(frac(1, 60))$] of a degree, and 1 second is #math.equation(block: false, alt: "the fraction 1 over 60")[$display(frac(1, 60))$] of a minute, or #math.equation(block: false, alt: "the fraction 1 over 3600")[$display(frac(1, 3600))$] of a degree. We use the following notation for minutes and seconds. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Fractions of a Degree] - One minute: #math.equation(block: false, alt: "1 prime equals the fraction 1 degrees over 60")[$" " " " " " " " " " " " 1^(′) = display(frac(1^(∘), 60))$] - One second: #math.equation(block: false, alt: "1 to the power ′ ′ equals the fraction 1 prime over 60 equals the fraction 1 degrees over 3600")[$" " " " " " " " " " " " 1^(′ ′) = display(frac(1^(′), 60)) = display(frac(1^(∘), 3600))$] ] When describing large distances, astronomers sometimes use the distance from Earth to the Sun, about 93 million miles, as the unit of measurement. This distance is called 1 #strong[Astronomical Unit], or 1 #strong[AU]. For example, an object that is three times as far away as the Sun would be at a distance of 3 AU. #examplebox("Example 7")[][ The star Wolf 359 has a parallax of #math.equation(block: false, alt: "0.85 to the power ′ ′")[$0.85^(′ ′)$] when observed from opposites sides of Earth's orbit. How far away is the star? #solutionbox[ In the figure below, the star Wolf 359 is located at point #math.equation(block: false, alt: "O")[$O$]. Our Sun is located at point #math.equation(block: false, alt: "C")[$C$], halfway between the two observation points at #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$]. Thus, the distance between the observation points is twice the distance from the Earth to the Sun, or 2 AU. #figure(figph[triangles for parallax], alt: "triangles for parallax", caption: none) The altitude from #math.equation(block: false, alt: "O")[$O$] to side #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$] forms a right triangle #math.equation(block: false, alt: "A C O")[$A C O$] and bisects the angle at #math.equation(block: false, alt: "O")[$O$]. Thus #math.equation(block: true, alt: "∠ A O C equals the fraction 1 over 2 open parenthesis 0.85 to the power ′ ′ close parenthesis equals 0.425 to the power ′ ′")[$upright(∠) A O C = display(frac(1, 2)) ( 0.85^(′ ′) ) = 0.425^(′ ′)$] We use the definition of tangent to find #math.equation(block: true, alt: "tan ⁡ 0.425 to the power ′ ′, equals the fraction opposite over adjacent; tan ⁡ open parenthesis 0.425 degrees / 3600 close parenthesis, equals the fraction 1 over x Solve for x .; x, equals the fraction 1 over tan ⁡ open parenthesis 0.425 degrees / 3600 close parenthesis approximately equals 485 , 000")[$tan 0.425^(′ ′) & = display(frac("opposite", "adjacent")) \ tan ( 0.425^(∘) / 3600 ) & = display(frac(1, x)) " " " " " " " " " " " " " " " " " " " " " " " " " " "Solve for " bold(italic(x)) . \ x & = display(frac(1, tan ( 0.425^(∘) / 3600 ))) ≈ 485 , 000$] The star is approximately #math.equation(block: false, alt: "485 , 000")[$485 , 000$] AU from Earth, or nearly half a million times the distance from Earth to the Sun. ] ] Two observers are 1 AU (astronomical unit) apart. They find that the parallax to a distant star is #math.equation(block: false, alt: "1 to the power ′ ′")[$1^(′ ′)$]. What is the distance to the star, in astronomical units? (This distance is called a #strong[parsec]. In other words, a parsec is the distance at which the parallax from observations 1 AU apart is #math.equation(block: false, alt: "1 to the power ′ ′")[$1^(′ ′)$]. In this Exercise, you are calculating the number of astronomical units in 1 parsec.) 1 parsec #math.equation(block: false, alt: "approximately equals 206 , 265")[$≈ 206 , 265$] AU Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Convert to a decimal fraction. + 5 inches #math.equation(block: false, alt: "equals")[$=$] \_\_\_\_\_\_ foot + 10 ounces #math.equation(block: false, alt: "equals")[$=$] \_\_\_\_\_\_ pound + 24 minutes #math.equation(block: false, alt: "equals")[$=$] \_\_\_\_\_\_ hour + 35 seconds #math.equation(block: false, alt: "equals")[$=$] \_\_\_\_\_\_ minute + 16 minutes #math.equation(block: false, alt: "equals")[$=$] \_\_\_\_\_\_ degree + #math.equation(block: false, alt: "4 prime equals")[$4^(′) =$] \_\_\_\_\_\_ #math.equation(block: false, alt: "∘")[$∘$] + #math.equation(block: false, alt: "2 to the power ′ ′ equals")[$2^(′ ′) =$] \_\_\_\_\_\_ #math.equation(block: false, alt: "∘")[$∘$] + #math.equation(block: false, alt: "1 prime 5 to the power ′ ′ equals")[$1^(′) " " 5^(′ ′) =$] \_\_\_\_\_\_#math.equation(block: false, alt: "∘")[$∘$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "0.416 bar")[$0.41 limits(6)^(―)$] + #math.equation(block: false, alt: "0.625")[$0.625$] + #math.equation(block: false, alt: "0.4")[$0.4$] + #math.equation(block: false, alt: "0.583 bar")[$0.58 limits(3)^(―)$] + #math.equation(block: false, alt: "0.26 bar")[$0.2 limits(6)^(―)$] + #math.equation(block: false, alt: "0.06 bar")[$0.0 limits(6)^(―)$] + #math.equation(block: false, alt: "0.0005 bar")[$0.000 limits(5)^(―)$] + #math.equation(block: false, alt: "0.01805 bar")[$0.0180 limits(5)^(―)$] ] === Section 3.2 Summary ==== Vocabulary - Parallax - Minute - Second ==== Concepts + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Law of Sines] 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: "the fraction sin ⁡ open parenthesis A close parenthesis over a equals the fraction sin ⁡ open parenthesis B close parenthesis over b equals the fraction sin ⁡ open parenthesis C close parenthesis over c")[$display(frac(sin ( A ), a)) = display(frac(sin ( B ), b)) = display(frac(sin ( C ), c))$]#figure(figph[triangle ABC], alt: "triangle ABC", caption: none) or equivalently, #math.equation(block: true, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis equals the fraction b over sin ⁡ open parenthesis B close parenthesis equals the fraction c over sin ⁡ open parenthesis C close parenthesis")[$display(frac(a, sin ( A ))) = display(frac(b, sin ( B ))) = display(frac(c, sin ( C )))$] ] + We can use the Law of Sines to find an unknown side in an oblique triangle. We must know the angle opposite the unknown side, and another side-angle pair. + We can also use the Law of Sines to find an unknown angle of a triangle. We must know two sides of the triangle and the angle opposite one of them. + Remember that there are two angles with a given sine. When using the Law of Sines, we must check whether both angles result in possible triangles. + We use minutes and seconds to measure very small angles. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Fractions of a Degree] - One minute: #math.equation(block: false, alt: "1 prime equals the fraction 1 degrees over 60")[$" " " " " " " " " " " " 1^(′) = display(frac(1^(∘), 60))$] - One second: #math.equation(block: false, alt: "1 to the power ′ ′ equals the fraction 1 prime over 60 equals the fraction 1 degrees over 3600")[$" " " " " " " " " " " " 1^(′ ′) = display(frac(1^(′), 60)) = display(frac(1^(∘), 3600))$] ] ==== Study Questions + Can we use the Law of Sines to solve a right triangle? + Explain why we cannot use the Law of Sines to solve the triangle with #math.equation(block: false, alt: "a equals 8 , b equals c")[$a = 8 , " " b = c$] and #math.equation(block: false, alt: "C equals 35 degrees")[$C = 35^(∘)$]. + Francine says "I'm thinking of an angle whose sine is #math.equation(block: false, alt: "0.3420")[$0.3420$] (rounded to four decimal places)." Delbert says "The angle must be #math.equation(block: false, alt: "20 degrees")[$20^(∘)$] (rounded to the nearest degree)." Is he correct? Why or why not? + Sketch two possible triangles with #math.equation(block: false, alt: "A equals 25 degrees , b equals 18")[$A = 25^(∘) , " " b = 18$], and #math.equation(block: false, alt: "a equals 10")[$a = 10$]. + Try to sketch a triangle with #math.equation(block: false, alt: "A equals 65 degrees , b equals 18")[$A = 65^(∘) , " " b = 18$], and #math.equation(block: false, alt: "a equals 10")[$a = 10$]. What went wrong? ==== Skills + Use the Law of Sines to find a side \#1-6 + Use the Law of Sines to find an angle \#7-12 + Use the Law of Sines to solve an oblique triangle \#13-18 + Solve problems using the Law of Sines \#19-28 + Compute distances using parallax \#29-32 + Solve problems involving the ambiguous case \#33-46 === Homework 3.2 For Problems 1–6, use the Law of Sines to find the indicated side. Round to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "x equals 7.85")[$x = 7.85$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "q equals 33.81")[$q = 33.81$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "d equals 28.37")[$d = 28.37$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 7–12, use the Law of Sines to find the indicated angle. Round to two decimal places. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "θ equals 30.80 degrees")[$θ = 30.80^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "θ equals 126.59 degrees")[$θ = 126.59^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "β equals 37.14 degrees")[$β = 37.14^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 13–18, sketch the triangle and solve. Round answers to two decimal places. #math.equation(block: true, alt: "b equals 7 , A equals 23 degrees , B equals 42 degrees")[$b = 7 , " " A = 23^(∘) , " " B = 42^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "a equals 4.09")[$a = 4.09$], #math.equation(block: true, alt: "c equals 9.48")[$" " c = 9.48$], #math.equation(block: true, alt: "C equals 115 degrees")[$" " C = 115^(∘)$] #math.equation(block: true, alt: "c equals 34 , A equals 53 degrees , C equals 26 degrees")[$c = 34 , " " A = 53^(∘) , " " C = 26^(∘)$] #math.equation(block: true, alt: "a equals 1.8 , c equals 2.1 , C equals 44 degrees")[$a = 1.8 , " " c = 2.1 , " " C = 44^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "b equals 2.98")[$b = 2.98$], #math.equation(block: true, alt: "A equals 36.54 degrees")[$" " A = 36.54^(∘)$], #math.equation(block: true, alt: "B equals 99.46 degrees")[$" " B = 99.46^(∘)$] #math.equation(block: true, alt: "b equals 8.5 , c equals 6.8 , B equals 23 degrees")[$b = 8.5 , " " c = 6.8 , " " B = 23^(∘)$] #math.equation(block: true, alt: "c equals 75 , A equals 35 degrees , B equals 46 degrees")[$c = 75 , " " A = 35^(∘) , " " B = 46^(∘)$] #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "a equals 43.55")[$a = 43.55$], #math.equation(block: true, alt: "b equals 54.62")[$" " b = 54.62$], #math.equation(block: true, alt: "C equals 99 degrees")[$" " C = 99^(∘)$] #math.equation(block: true, alt: "a equals 94 , B equals 29 degrees , C equals 84 degrees")[$a = 94 , " " B = 29^(∘) , " " C = 84^(∘)$] For Problems 19–26, + Sketch and label a triangle to illustrate the problem. + Solve the problem. Maryam wants to know the height of a cliff on the other side of a ravine. The angle of elevation from her edge of the ravine to the cliff top is #math.equation(block: false, alt: "84.6 degrees")[$84.6^(∘)$]. When she moves 30 feet back from the ravine, the angle of elevation is #math.equation(block: false, alt: "82.5 degrees")[$82.5^(∘)$]. How tall is the cliff? a. #figure(figph[triangle], alt: "triangle", caption: none) b. 808.1 ft Amir wants to know the height of a tree in the median strip of a highway. The angle of elevation from the highway shoulder to the treetop is #math.equation(block: false, alt: "43.5 degrees")[$43.5^(∘)$]. When he moves 10 feet farther away from the tree, the angle of elevation is #math.equation(block: false, alt: "37.2 degrees")[$37.2^(∘)$]. How tall is the tree? Delbert and Francine are 10 kilometers apart, both observing a satellite that passes directly over their heads. At a moment when the satellite is between them, Francine measures its angle of elevation as #math.equation(block: false, alt: "84.6 degrees")[$84.6^(∘)$], and Delbert measures an angle of #math.equation(block: false, alt: "87 degrees")[$87^(∘)$]. How far is the satellite from Delbert? a. #figure(figph[triangle], alt: "triangle", caption: none) b. 68.2 km Megan rows her kayak due east. When she began, she spotted a lighthouse 2000 meters in the distance at an angle of #math.equation(block: false, alt: "14 degrees")[$14^(∘)$] south of east. After traveling for of an hour, the lighthouse was at an angle of #math.equation(block: false, alt: "83 degrees")[$83^(∘)$] south of east. How far did Megan travel, and what was her average speed? Chad is hiking along a straight path but needs to detour around a large pond. He turns #math.equation(block: false, alt: "23 degrees")[$23^(∘)$] from his path until clear of the pond, then walks back to his original path, intercepting it at an angle of #math.equation(block: false, alt: "29 degrees")[$29^(∘)$] and at a distance of 2 miles from where he had left the path. How far did Chad walk in each of the two segments of his detour, and how much farther did his detour require compared with a straight line through the pond? #figure(figph[pond], alt: "pond", caption: none) a. #figure(figph[triangle], alt: "triangle", caption: none) b.1.23 mi #math.equation(block: false, alt: "plus")[$+$] 0.99 mi; 0.22 mi Bob is flying to Monterey but must change course to avoid a storm. He flies #math.equation(block: false, alt: "19 degrees")[$19^(∘)$] off from his original direction until he clears the storm, then turns again to return back to his original flight path, intercepting it at an angle of #math.equation(block: false, alt: "54.9 degrees")[$54.9^(∘)$] and at a distance of 50 miles from where he had left it. How much farther did his detour require compared with his original course? #figure(figph[flight path change], alt: "flight path change", caption: none) Geologists find an outcropping from an underground rock formation that normally indicates the presence of oil. The outcropping is on a hillside, and the formation itself dips another #math.equation(block: false, alt: "17 degrees")[$17^(∘)$] from the surface. If an oil well is placed 1000 meters downhill from the outcropping, how far will the well have to drill before it reaches the formation? #figure(figph[oil], alt: "oil", caption: none) a. #figure(figph[triangle], alt: "triangle", caption: none) b. 322.6 m A proposed ski lift will rise from point near the base of the slope with an angle of #math.equation(block: false, alt: "27 degrees")[$27^(∘)$]. At a distance of 400 meters further from the slope, the angle of elevation to the top of the ski lift is #math.equation(block: false, alt: "19 degrees")[$19^(∘)$]. How long is the ski lift? #figure(figph[ski lift], alt: "ski lift", caption: none) Thelma wants to measure the height of a hill. She first plants a 50 foot tall antenna at the hill's peak. Then she descends the hill and finds a point where she can see the top and the bottom of the antenna. The angle of elevation to the bottom of the antenna is #math.equation(block: false, alt: "23 degrees")[$23^(∘)$], and the angle of elevation to the top of the antenna is #math.equation(block: false, alt: "24 degrees")[$24^(∘)$]. #figure(figph[hill], alt: "hill", caption: none) + Find #math.equation(block: false, alt: "∠ A C B")[$upright(∠) A C B$]. + Find#math.equation(block: false, alt: "∠ C A B")[$upright(∠) C A B$], at the top of the antenna. + How long is #math.equation(block: false, alt: "B C")[$B C$], the distance from the bottom of the antenna to #math.equation(block: false, alt: "C")[$C$]? + How tall is the hill? + #math.equation(block: false, alt: "1 degrees")[$1^(∘)$] + #math.equation(block: false, alt: "66 degrees")[$66^(∘)$] + 2617.2 ft + 1022.6 ft A billboard of California's gubernatorial candidate Angelyne is located on the roof of a building. At a distance of 180 feet from the building, the angles of elevation to the bottom and top of the billboard are respectively #math.equation(block: false, alt: "39.8 degrees")[$39.8^(∘)$] and #math.equation(block: false, alt: "47.3 degrees")[$47.3^(∘)$]. How tall is the billboard? #figure(figph[billboard], alt: "billboard", caption: none) For Problems 29–32, compute the following distances in Astronomical Units. Then convert to kilometers, using the fact that 1 AU #math.equation(block: false, alt: "approximately equals 1.5 times 10 to the power 8")[$≈ 1.5 × 10^(8)$] km. When observed from opposite sides of Earth's orbit, the star Alpha Centauri has a parallax of #math.equation(block: false, alt: "0.76 to the power ′ ′")[$0.76^(′ ′)$]. How far from the Sun is Alpha Centauri? 540,000 AU #math.equation(block: false, alt: "approximately equals 8.1 times 10 to the power 13")[$≈ 8.1 × 10^(13)$] km How far from the Sun is Barnard's star, which has a parallax of #math.equation(block: false, alt: "1.1 to the power ′ ′")[$1.1^(′ ′)$] when observed at opposite ends of Earth's orbit? How far from the Sun is Tau Ceti, which has a parallax of #math.equation(block: false, alt: "0.55 to the power ′ ′")[$0.55^(′ ′)$] when observed from opposite ends of Earth's orbit? 750,000 AU #math.equation(block: false, alt: "approximately equals 1.1 times 10 to the power 14")[$≈ 1.1 × 10^(14)$] km How far from the Sun is Sirius, which has a parallax of #math.equation(block: false, alt: "0.75 to the power ′ ′")[$0.75^(′ ′)$] when observed from opposite ends of Earth's orbit? Problems 33–38 consider the ambiguous case of the Law of Sines, when two sides and an angle opposite one of them are known. In the right triangle #math.equation(block: false, alt: "A B C")[$A B C$] shown, #math.equation(block: false, alt: "∠ A equals 30 degrees , ∠ C equals 90 degrees")[$upright(∠) A = 30^(∘) , " " upright(∠) C = 90^(∘)$], and #math.equation(block: false, alt: "c equals 3")[$c = 3$] inches. #figure(figph[triangle], alt: "triangle", caption: none) + Use the definition of #math.equation(block: false, alt: "sin ⁡ open parenthesis A close parenthesis")[$sin ( A )$] to solve for #math.equation(block: false, alt: "a")[$a$] (the length of side #math.equation(block: false, alt: "B C bar")[$limits(B C)^(―)$]). + Can you draw a triangle #math.equation(block: false, alt: "A B C")[$A B C$] with #math.equation(block: false, alt: "A equals 30 degrees")[$A = 30^(∘)$] and #math.equation(block: false, alt: "c equals 3")[$c = 3$] if #math.equation(block: false, alt: "a less than the fraction 3 over 2")[$a < display(frac(3, 2))$]? Why or why not? + How many triangles are possible if #math.equation(block: false, alt: "the fraction 3 over 2 less than a less than 3")[$display(frac(3, 2)) < a < 3$]? + How many triangles are possible if #math.equation(block: false, alt: "a greater than 3")[$a > 3$]? + #math.equation(block: false, alt: "the fraction 3 over 2")[$display(frac(3, 2))$] + No, #math.equation(block: false, alt: "a")[$a$] is too short. + 2 + 1 In this problem we show that there are two different triangles #math.equation(block: false, alt: "A B C")[$A B C$] with #math.equation(block: false, alt: "A equals 30 degrees , a equals 2")[$A = 30^(∘) , " " a = 2$] and #math.equation(block: false, alt: "c equals 3")[$c = 3$]. + Use a protractor to draw an angle #math.equation(block: false, alt: "A equals 30 degrees")[$A = 30^(∘)$]. Mark point #math.equation(block: false, alt: "B")[$B$] on one side of the angle so that #math.equation(block: false, alt: "A B bar")[$limits(A B)^(―)$] is 3 inches long. + Locate two distinct points on the other side of the angle that are each 2 inches from point #math.equation(block: false, alt: "B")[$B$]. These points are both possible locations for point #math.equation(block: false, alt: "C")[$C$]. + Use the Law of Sines to find two distinct possible measures for #math.equation(block: false, alt: "∠ C")[$upright(∠) C$]. In #math.equation(block: false, alt: "△ A B C , ∠ A equals 30 degrees")[$upright(△) A B C , upright(∠) A = 30^(∘)$] and #math.equation(block: false, alt: "c equals 12")[$c = 12$]. How many triangles are possible for each of the following lengths for side #math.equation(block: false, alt: "a")[$a$]? Sketch the solutions in each case. + #math.equation(block: false, alt: "a equals 6")[$a = 6$] + #math.equation(block: false, alt: "a equals 4")[$a = 4$] + #math.equation(block: false, alt: "a equals 9")[$a = 9$] + #math.equation(block: false, alt: "a equals 15")[$a = 15$] + 1, #figure(figph[triangle], alt: "triangle", caption: none) + 0, #figure(figph[triangle], alt: "triangle", caption: none) + 2, #figure(figph[triangle], alt: "triangle", caption: none) + 1, #figure(figph[triangle], alt: "triangle", caption: none) Consider the triangle #math.equation(block: false, alt: "A B C")[$A B C$] shown below. #figure(figph[triangle], alt: "triangle", caption: none) + Express the length of the altitude in terms of #math.equation(block: false, alt: "∠ A")[$upright(∠) A$] and #math.equation(block: false, alt: "c")[$c$]. + Now suppose we keep #math.equation(block: false, alt: "∠ A")[$upright(∠) A$] and side #math.equation(block: false, alt: "c")[$c$] fixed, but allow #math.equation(block: false, alt: "a")[$a$] to vary in length. What is the smallest value #math.equation(block: false, alt: "a")[$a$] can have and still be long enough to make a triangle? + What are the largest and smallest values that #math.equation(block: false, alt: "a")[$a$] can have in order to produce two distinct triangles #math.equation(block: false, alt: "A B C")[$A B C$] (without changing #math.equation(block: false, alt: "∠ A")[$upright(∠) A$] and side #math.equation(block: false, alt: "c")[$c$])? For the triangle in Problem 36, suppose #math.equation(block: false, alt: "A equals 40 degrees")[$A = 40^(∘)$] and #math.equation(block: false, alt: "c equals 8")[$c = 8$]. + Sketch and solve the triangle if #math.equation(block: false, alt: "a equals 12")[$a = 12$]. + Sketch and solve the triangle if #math.equation(block: false, alt: "a equals 6")[$a = 6$]. + Sketch and solve the triangle if #math.equation(block: false, alt: "a equals 4")[$a = 4$]. + For what value of #math.equation(block: false, alt: "a")[$a$] is #math.equation(block: false, alt: "c")[$c$] the hypotenuse of a right triangle? + #math.equation(block: false, alt: "C equals 25.37 degrees , B equals 114.63 degrees , b equals 16.97")[$C = 25.37^(∘) , " " B = 114.63^(∘) , " " b = 16.97$] + #math.equation(block: false, alt: "C equals 58.99 degrees , B equals 81.01 degrees , b equals 9.22")[$C = 58.99^(∘) , " " B = 81.01^(∘) , " " b = 9.22$] or #math.equation(block: false, alt: "C equals 121.01 degrees , B equals 18.99 degrees , b equals 3.04")[$C = 121.01^(∘) , " " B = 18.99^(∘) , " " b = 3.04$] + no solution + 5.14 For the figure in Problem 36, suppose #math.equation(block: false, alt: "A equals 70 degrees")[$A = 70^(∘)$] and #math.equation(block: false, alt: "c equals 20")[$c = 20$]. + For what value of #math.equation(block: false, alt: "a")[$a$] is the triangle a right triangle? + For what values of #math.equation(block: false, alt: "a")[$a$] are there two solutions for the triangle? + For what values of #math.equation(block: false, alt: "a")[$a$] is there one obtuse solution for the triangle? + For what value of #math.equation(block: false, alt: "a")[$a$] is there no solution? For Problems 39–42, find the remaining angles of the triangle. Round answers to two decimal places. (These problems involve the ambiguous case.) #math.equation(block: true, alt: "a equals 66 , c equals 43 , C equals 25 degrees")[$a = 66 , " " c = 43 , " " C = 25^(∘)$] #math.equation(block: false, alt: "A equals 40.44 degrees , B equals 114.56 degrees")[$A = 40.44^(∘) , " " B = 114.56^(∘)$] or #math.equation(block: false, alt: "A equals 139.56 degrees , B equals 15.44 degrees")[$A = 139.56^(∘) , " " B = 15.44^(∘)$] #math.equation(block: true, alt: "b equals 10 , c equals 14 , B equals 20 degrees")[$b = 10 , " " c = 14 , " " B = 20^(∘)$] #math.equation(block: true, alt: "b equals 100 , c equals 80 , B equals 49 degrees")[$b = 100 , " " c = 80 , " " B = 49^(∘)$] #math.equation(block: true, alt: "C equals 37.14 degrees , A equals 93.86 degrees")[$C = 37.14^(∘) , " " A = 93.86^(∘)$] #math.equation(block: true, alt: "b equals 4.7 , c equals 6.3 , C equals 54 degrees")[$b = 4.7 , " " c = 6.3 , " " C = 54^(∘)$] Delbert and Francine are 1000 yards apart. The angle Delbert sees between Francine and a certain tree is #math.equation(block: false, alt: "38 degrees")[$38^(∘)$]. If the tree is 800 yards from Francine, how far is it from Delbert? (There are two possible answers.) 1299 yd or 277.2 yd From the lookout point on Fabrick Rock, Ann can see not only see the famous "Crooked Spire" in Chesterfield, which is 8 miles away, but also the red phone box in the village of Alton. Chesterfield and Alton are 7 miles apart. Fabrick Rock has a plaque that shows directions to famous sites, and from the plaque Ann determines that the angle between the lines to the spire and the phone box measures #math.equation(block: false, alt: "19 degrees")[$19^(∘)$]. How far is Fabrick Rock from the phone box? (There are two possible answers.) + Sketch a triangle with #math.equation(block: false, alt: "A equals 25 degrees , B equals 35 degrees")[$A = 25^(∘) , " " B = 35^(∘)$], and #math.equation(block: false, alt: "b equals 16")[$b = 16$]. + Use the Law of Sines to find #math.equation(block: false, alt: "a")[$a$]. + Use the Law of Sines to find #math.equation(block: false, alt: "c")[$c$]. + Find #math.equation(block: false, alt: "c")[$c$] without using the Law of Sines. (Hint: Sketch the altitude, #math.equation(block: false, alt: "h")[$h$], from #math.equation(block: false, alt: "C")[$C$] to make two right triangles. Find #math.equation(block: false, alt: "h")[$h$], then use #math.equation(block: false, alt: "h")[$h$] to find #math.equation(block: false, alt: "c")[$c$].) + 11.79 + 24.16 + 24.16 + Sketch a triangle with #math.equation(block: false, alt: "A equals 75 degrees , a equals 15")[$A = 75^(∘) , " " a = 15$], and #math.equation(block: false, alt: "b equals 6")[$b = 6$]. + Use the Law of Sines to find #math.equation(block: false, alt: "c")[$c$]. + Find #math.equation(block: false, alt: "c")[$c$] without using the Law of Sines. Problems 47–48 prove the Law of Sines using the formula for the area of a triangle. (See Section 3.1 for the appropriate formula.) Sketch a triangle with angles #math.equation(block: false, alt: "A , B")[$A , " " B$] and #math.equation(block: false, alt: "C")[$C$] and opposite sides of lengths respectively #math.equation(block: false, alt: "a , b")[$a , " " b$] and #math.equation(block: false, alt: "c")[$c$]. + Write the area of the triangle in terms of #math.equation(block: false, alt: "a , b")[$a , " " b$], and angle #math.equation(block: false, alt: "C")[$C$]. + Write the area of the triangle in terms of #math.equation(block: false, alt: "a , c")[$a , " " c$], and angle #math.equation(block: false, alt: "B")[$B$]. + Write the area of the triangle in terms of #math.equation(block: false, alt: "b , c")[$b , " " c$], and angle #math.equation(block: false, alt: "A")[$A$]. #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "the fraction 1 over 2 a b sin ⁡ open parenthesis C close parenthesis")[$display(frac(1, 2)) a b sin ( C )$] + #math.equation(block: false, alt: "the fraction 1 over 2 a c sin ⁡ open parenthesis B close parenthesis")[$display(frac(1, 2)) a c sin ( B )$] + #math.equation(block: false, alt: "the fraction 1 over 2 b c sin ⁡ open parenthesis A close parenthesis")[$display(frac(1, 2)) b c sin ( A )$] Equate the three different expressions from Problem 47 for the area of the triangle. Multiply through by #math.equation(block: false, alt: "the fraction 2 over a b c")[$display(frac(2, a b c))$] and simplify to deduce the Law of Sines. Here is a method for solving certain oblique triangles by dividing them into two right triangles. In the triangle shown, we know two angles, #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], and the side opposite one of them, say #math.equation(block: false, alt: "a")[$a$]. We would like to find side #math.equation(block: false, alt: "b")[$b$]. #figure(figph[triangle], alt: "triangle", caption: none) + Draw the altitude #math.equation(block: false, alt: "h")[$h$] from angle #math.equation(block: false, alt: "C")[$C$]. + Write an expression for #math.equation(block: false, alt: "b")[$b$] in terms of #math.equation(block: false, alt: "h")[$h$] and angle #math.equation(block: false, alt: "A")[$A$]. + Write an expression for #math.equation(block: false, alt: "h")[$h$] in terms of angle #math.equation(block: false, alt: "B")[$B$]. + Substitute your expression for #math.equation(block: false, alt: "h")[$h$] into your expression for #math.equation(block: false, alt: "b")[$b$]. + Which of the following is equivalent to the formula you wrote in part (d)?+ #math.equation(block: false, alt: "a sin ⁡ open parenthesis A close parenthesis equals b sin ⁡ open parenthesis B close parenthesis")[$a sin ( A ) = b sin ( B )$] + #math.equation(block: false, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis equals the fraction b over sin ⁡ open parenthesis B close parenthesis")[$display(frac(a, sin ( A ))) = display(frac(b, sin ( B )))$] + #math.equation(block: false, alt: "the fraction a over sin ⁡ open parenthesis B close parenthesis equals the fraction b over sin ⁡ open parenthesis A close parenthesis")[$display(frac(a, sin ( B ))) = display(frac(b, sin ( A )))$] + #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "b equals the fraction h over sin ⁡ open parenthesis A close parenthesis")[$b = display(frac(h, sin ( A )))$] + #math.equation(block: false, alt: "h equals a sin ⁡ open parenthesis B close parenthesis")[$h = a sin ( B )$] + #math.equation(block: false, alt: "b equals the fraction a sin ⁡ open parenthesis B close parenthesis over sin ⁡ open parenthesis A close parenthesis")[$b = display(frac(a sin ( B ), sin ( A )))$] + ii