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== 3.4#h(0.6em)Chapter 3 Summary and Review


    
      
=== Key Concepts


      + We put an angle #math.equation(block: false, alt: "θ")[$θ$] in #emph[standard position] by placing its vertex at the origin and the #emph[initial side] on the positive #math.equation(block: false, alt: "x")[$x$]-axis.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Coordinate Definitions of the Trigonometric Ratios]

- #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r")[$cos ( θ ) = display(frac(x, r))$]
- #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r")[$sin ( θ ) = display(frac(y, r))$]
- #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x")[$tan ( θ ) = display(frac(y, x))$]

#figure(figph[supplementary angles in standard position], alt: "supplementary angles in standard position", caption: none)
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Trigonometric Ratios for Supplementary Angles]

- #math.equation(block: false, alt: "cos ⁡ open parenthesis 180 degrees minus θ close parenthesis equals minus cos ⁡ open parenthesis θ close parenthesis")[$cos ( 180^(∘) − θ ) = − cos ( θ )$]
- #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$sin ( 180^(∘) − θ ) = sin ( θ )$]
- #math.equation(block: false, alt: "tan ⁡ open parenthesis 180 degrees minus θ close parenthesis equals minus tan ⁡ open parenthesis θ close parenthesis")[$tan ( 180^(∘) − θ ) = − tan ( θ )$]

#figure(figph[supplementary angles in standard position], alt: "supplementary angles in standard position", caption: none)
]
+ There are always two (supplementary) angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] that have the same sine.  Your calculator will only tell you one of them.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Area of a Triangle]

If a triangle has sides of length #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], and the angle between those two sides is #math.equation(block: false, alt: "θ")[$θ$], then the area of the triangle is given by

#math.equation(block: true, alt: "A equals the fraction 1 over 2 a b sin ⁡ open parenthesis θ close parenthesis")[$A = display(frac(1, 2)) a b sin ( θ )$]#figure(figph[Triangle-angle-theta-sides-a-and-b], alt: "Triangle-angle-theta-sides-a-and-b", caption: none)
]
+ #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")[$" " " " " " " " " " " " 1^(′ ′) = display(frac(1^(′), 60)) = display(frac(1^(∘), 3600))$]
]
+ You can remember the trig values for the special angles if you memorize two triangles:#figure(figph[45-45-90 triangle], alt: "45-45-90 triangle", caption: none)

#figure(figph[30-60-90 triangle], alt: "30-60-90 triangle", caption: none)
+ For the trigonometric ratios of most angles, your calculator gives approximations, not exact values.
+ 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],
))
]


    
    
      
=== Chapter 3 Review Problems


      Use facts about supplementary angles to answer the questions in Problems 1 and 2.


      
        If #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 2")[$sin ( θ ) = display(frac(1, 2))$], what is #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$sin ( 180^(∘) − θ )$]? What are the possible values for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]?


        
          #math.equation(block: true, alt: "the fraction 1 over 2 , the fraction plus or minus the square root of 3 over 2")[$display(frac(1, 2)) , " " display(frac(± sqrt(3), 2))$]


        
      
      
        If #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5")[$sin ( θ ) = display(frac(3, 5))$], what is #math.equation(block: false, alt: "sin ⁡ open parenthesis 180 degrees minus θ close parenthesis")[$sin ( 180^(∘) − θ )$]? What are the possible values for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]?


      
      
        Two sides of a triangle are 12 and 9 units long. The angle #math.equation(block: false, alt: "θ")[$θ$] between those sides is #math.equation(block: false, alt: "66 degrees")[$66^(∘)$].


        + Sketch the triangle with #math.equation(block: false, alt: "θ")[$θ$] in standard position.
+ What is the area of the triangle?
+ Draw another triangle with the same area you found in (b) and with sides 12 and 9, but with an obtuse angle between those two sides. What is the obtuse angle?


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ 49.33
+ #figure(figph[triangle], alt: "triangle", caption: none)

 #linebreak() #math.equation(block: false, alt: "114 degrees")[$114^(∘)$]


        
      
      
        A triangle has base 5 units and altitude 6 units, and another side of length 8 units.


        + Sketch the triangle so that the acute angle #math.equation(block: false, alt: "θ")[$θ$] between the sides of length 5 and 8 is in standard position.
+ What is the area of the triangle?
+ What is #math.equation(block: false, alt: "θ")[$θ$] ?
+ Draw another triangle with the same area you found in (b) and with sides 5 and 8, but with an obtuse angle between those two sides. What is the obtuse angle?


      
      For Problems 5–12,


      + Sketch an angle in standard position with the given properties.
+ Find #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis ,")[$cos ( θ ) , " " sin ( θ ) ,$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$].
+ Find the angle #math.equation(block: false, alt: "θ")[$θ$], rounded to tenths of a degree.


      
        The point #math.equation(block: false, alt: "open parenthesis minus 2 , 3 close parenthesis")[$( − 2 , 3 )$] is on the terminal side.


        #figure(figph[grid], alt: "grid", caption: none)


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 2 over the square root of 13")[$cos ( θ ) = display(frac(− 2, sqrt(13)))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 3 over the square root of 13")[$" " sin ( θ ) = display(frac(3, sqrt(13)))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 2")[$" " tan ( θ ) = display(frac(− 3, 2))$]
+ #math.equation(block: false, alt: "θ equals 123.7 degrees")[$θ = 123.7^(∘)$]


        
      
      
        The point #math.equation(block: false, alt: "open parenthesis minus 1 , 6 close parenthesis")[$( − 1 , 6 )$] is on the terminal side.


        #figure(figph[grid], alt: "grid", caption: none)


      
      
        The point on the terminal side 20 units from the origin has #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "minus 16")[$− 16$].


        #figure(figph[grid], alt: "grid", caption: none)


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 4 over 5")[$cos ( θ ) = display(frac(− 4, 5))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5")[$" " sin ( θ ) = display(frac(3, 5))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 4")[$" " tan ( θ ) = display(frac(− 3, 4))$]
+ #math.equation(block: false, alt: "θ equals 143.1 degrees")[$θ = 143.1^(∘)$]


        
      
      
        The point on the terminal side 25 units from the origin has #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "minus 7")[$− 7$].


        #figure(figph[grid], alt: "grid", caption: none)


      
      
        #math.equation(block: false, alt: "θ")[$θ$] is obtuse, and the point on the terminal side 6 units from the origin has #math.equation(block: false, alt: "y")[$y$]-coordinate #math.equation(block: false, alt: "5")[$5$].


        #figure(figph[grid], alt: "grid", caption: none)


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 11 over 6")[$cos ( θ ) = display(frac(− sqrt(11), 6))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 6")[$" " sin ( θ ) = display(frac(5, 6))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over the square root of 11")[$" " tan ( θ ) = display(frac(− 5, sqrt(11)))$]
+ #math.equation(block: false, alt: "θ equals 123.6 degrees")[$θ = 123.6^(∘)$]


        
      
      
        #math.equation(block: false, alt: "θ")[$θ$] is obtuse, and the point on the terminal side 4 units from the origin has #math.equation(block: false, alt: "y")[$y$]-coordinate #math.equation(block: false, alt: "1")[$1$].


        #figure(figph[grid], alt: "grid", caption: none)


      
      
        #math.equation(block: false, alt: "θ")[$θ$] is obtuse, and #math.equation(block: false, alt: "sin ⁡ θ equals 0.96.")[$sin θ = 0.96 .$]


        #figure(figph[grid], alt: "grid", caption: none)


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 7 over 25")[$cos ( θ ) = display(frac(− 7, 25))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 24 over 25")[$" " sin ( θ ) = display(frac(24, 25))$], #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 24 over 7")[$" " tan ( θ ) = display(frac(− 24, 7))$]
+ #math.equation(block: false, alt: "θ equals 106.3 degrees")[$θ = 106.3^(∘)$]


        
      
      
        #math.equation(block: false, alt: "θ")[$θ$] is obtuse, and #math.equation(block: false, alt: "sin ⁡ θ equals 0.8.")[$sin θ = 0.8 .$]


        #figure(figph[grid], alt: "grid", caption: none)


      
      For Problems 13–16, solve the equation. Round to the nearest #math.equation(block: false, alt: "0.1 degrees")[$0.1^(∘)$].


      
        #math.equation(block: true, alt: "2.5 sin ⁡ open parenthesis θ close parenthesis equals 0.43")[$2.5 sin ( θ ) = 0.43$]


        
          #math.equation(block: true, alt: "9.9 degrees")[$9.9^(∘)$], #math.equation(block: true, alt: "170.1 degrees")[$" " 170.1^(∘)$]


        
      
      
        #math.equation(block: true, alt: "5 sin ⁡ open parenthesis θ close parenthesis equals 1.25")[$5 sin ( θ ) = 1.25$]


      
      
        #math.equation(block: true, alt: "the fraction 3 over sin ⁡ open parenthesis θ close parenthesis equals 8")[$display(frac(3, sin ( θ ))) = 8$]


        
          #math.equation(block: true, alt: "22.0 degrees , 158.0 degrees")[$22.0^(∘) , " " 158.0^(∘)$]


        
      
      
        #math.equation(block: true, alt: "0.6 equals the fraction sin ⁡ open parenthesis θ close parenthesis over 1.5 equals 8")[$0.6 = display(frac(sin ( θ ), 1.5)) = 8$]


      
      For Problems 17–18,


      + Find exact values for the base and height of the triangle.
+ Compute an exact value for the area of the triangle.


      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          + #math.equation(block: false, alt: "7 the square root of 2")[$7 sqrt(2)$]
+ #math.equation(block: false, alt: "28 the square root of 2")[$28 sqrt(2)$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      
        Find the area of the triangular plot of land shown at right if #math.equation(block: false, alt: "θ equals 109.9 degrees")[$θ = 109.9^(∘)$].


        #figure(figph[triangle], alt: "triangle", caption: none)


        
          5127.39 sq ft


        
      
      
        What is the area of the triangular piece of tile shown at right?


        #figure(figph[triangle], alt: "triangle", caption: none)


      
      For Problems 21–22, 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: "20.41 degrees")[$20.41^(∘)$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      For Problems 23–26, 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: "a equals 27.86")[$a = 27.86$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          #math.equation(block: true, alt: "b equals 6.03")[$b = 6.03$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      For Problems 27–30, use the Law of Cosines to find the indicated side. Round to two decimal places.


      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          #math.equation(block: true, alt: "w equals 62.10")[$w = 62.10$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


        
          #math.equation(block: true, alt: "s equals 15.61 or 57.45")[$s = 15.61 " " "or" " " 57.45$]


        
      
      
        #figure(figph[triangle], alt: "triangle", caption: none)


      
      For Problems 31–40,


      + Sketch the triangle described.
+ Find the missing angle or length. Round answers to two decimal places.


      
        #math.equation(block: true, alt: "A equals 111 degrees")[$A = 111^(∘)$], #math.equation(block: true, alt: "b equals 9")[$" " b = 9$], #math.equation(block: true, alt: "c equals 4")[$" " c = 4$], #math.equation(block: true, alt: "a equals ?")[$" " a = ?$]


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ 8.82


        
      
      
        #math.equation(block: false, alt: "B equals 32 degrees")[$B = 32^(∘)$], #math.equation(block: false, alt: "a equals 7")[$" " a = 7$], #math.equation(block: false, alt: "c equals 5")[$" " c = 5$], #math.equation(block: false, alt: "b equals")[$" " b =$] ?


      
      
        #math.equation(block: true, alt: "C equals 87 degrees")[$C = 87^(∘)$], #math.equation(block: true, alt: "a equals 26")[$" " a = 26$], #math.equation(block: true, alt: "b equals 42")[$" " b = 42$], #math.equation(block: true, alt: "A equals ?")[$" " A = ?$]


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ #math.equation(block: false, alt: "32.57 degrees")[$32.57^(∘)$]


        
      
      
        #math.equation(block: true, alt: "C equals 11 degrees")[$C = 11^(∘)$], #math.equation(block: true, alt: "a equals 49")[$" " a = 49$], #math.equation(block: true, alt: "b equals 56")[$" " b = 56$], #math.equation(block: true, alt: "B equals ?")[$" " B = ?$]


      
      
        #math.equation(block: true, alt: "a equals 16")[$a = 16$], #math.equation(block: true, alt: "B equals 88 degrees")[$" " B = 88^(∘)$], #math.equation(block: true, alt: "C equals 13 degrees")[$" " C = 13^(∘)$], #math.equation(block: true, alt: "b equals ?")[$" " b = ?$]


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ 16.29


        
      
      
        #math.equation(block: true, alt: "b equals 71")[$b = 71$], #math.equation(block: true, alt: "A equals 19 degrees")[$" " A = 19^(∘)$], #math.equation(block: true, alt: "C equals 110 degrees")[$" " C = 110^(∘)$], #math.equation(block: true, alt: "a equals ?")[$" " a = ?$]


      
      
        #math.equation(block: true, alt: "a equals 5.8")[$a = 5.8$], #math.equation(block: true, alt: "b equals 4.6")[$" " b = 4.6$], #math.equation(block: true, alt: "c equals 5.2")[$" " c = 5.2$], #math.equation(block: true, alt: "C equals ?")[$" " C = ?$]


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ #math.equation(block: false, alt: "58.65 degrees")[$58.65^(∘)$]


        
      
      
        #math.equation(block: true, alt: "a equals 17.7")[$a = 17.7$], #math.equation(block: true, alt: "b equals 18.4")[$" " b = 18.4$], #math.equation(block: true, alt: "c equals 28.1")[$" " c = 28.1$], #math.equation(block: true, alt: "A equals ?")[$" " A = ?$]


      
      
        #math.equation(block: true, alt: "a equals 52")[$a = 52$], #math.equation(block: true, alt: "b equals 36")[$" " b = 36$], #math.equation(block: true, alt: "B equals 19 degrees")[$" " B = 19^(∘)$], #math.equation(block: true, alt: "c equals ?")[$" " c = ?$]


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ 17.40


          or


          + #figure(figph[triangle], alt: "triangle", caption: none)
+ 80.93


        
      
      
        #math.equation(block: true, alt: "a equals 83")[$a = 83$], #math.equation(block: true, alt: "c equals 133")[$" " c = 133$], #math.equation(block: true, alt: "A equals 31 degrees")[$" " A = 31^(∘)$], #math.equation(block: true, alt: "b equals ?")[$" " b = ?$]


      
      For Problems 41–46,


      + Sketch and label a triangle to illustrate the problem.
+ Use the Law of Sines and/or the Law of Cosines to answer the questions.


      
        A radio tower is 40 miles from an airport, in the direction of #math.equation(block: false, alt: "23 degrees")[$23^(∘)$] east of north. Maria flies due north from the airport, and after 12 minutes, she sees the radio tower in the direction #math.equation(block: false, alt: "37 degrees")[$37^(∘)$] east of north. How far has she traveled, and what is her average ground speed?


        
          + #figure(figph[triangle], alt: "triangle", caption: none)
+ 16.08 mi, 80.4 mph


        
      
      
        As Odysseus begins a sailing journey, an island 2 kilometers distant is in the direction #math.equation(block: false, alt: "12 degrees")[$12^(∘)$] west of north. After he sails due east for a time, the island is #math.equation(block: false, alt: "75 degrees")[$75^(∘)$] west of north. How far has Odysseus sailed?


      
      
        Delbert and Francine are standing 100 meters apart on one side of a stream. A tree lies on the opposite shore. The angle Delbert sees from Francine to the tree is #math.equation(block: false, alt: "37 degrees")[$37^(∘)$], and the angle Francine sees from Delbert to the tree is #math.equation(block: false, alt: "46 degrees")[$46^(∘)$]. How far is Delbert from tree?


        
          + #figure(figph[triangle Francine-Delbert-Tree], alt: "triangle Francine-Delbert-Tree", caption: none)
+ 72.47


        
      
      
        Evel and Carla are estimating the distance across a canyon. They stand 75 meters apart and each site a larger boulder on the opposite side. Evel measures an angle of #math.equation(block: false, alt: "82 degrees")[$82^(∘)$] from Carla to the boulder, and Carla measures an angle of #math.equation(block: false, alt: "74 degrees")[$74^(∘)$] from Evel to the boulder.  How far is Evel from the boulder?


      
      
        A blimp is flying in a straight line towards a football stadium. Giselle and Hakim have homes 520 meters apart, both directly below the blimp's path. At a moment when the blimp is between them, Giselle measures an angle of elevation to the blimp to be #math.equation(block: false, alt: "38 degrees")[$38^(∘)$], and Hakim measures the angle of elevation to be #math.equation(block: false, alt: "42 degrees")[$42^(∘)$].


        + How far is the blimp from Giselle?
+ How high is the blimp above the ground?


        
          + #figure(figph[triangle Giselle-Hakim-blimp], alt: "triangle Giselle-Hakim-blimp", caption: none)

 #linebreak() 353.32
+ 217.52 m


        
      
      
        A model plane flying in a straight line passes directly over both first Adi's and then Bettina's head. Adi sees an angle of elevation of #math.equation(block: false, alt: "17 degrees")[$17^(∘)$] to the plane and Bettina sees an angle of elevation of #math.equation(block: false, alt: "58 degrees")[$58^(∘)$]. Adi and Bettina are 150 meters apart, and Bettina is between Adi and the plane.


        + How far is the model plane from Bettina?
+ How high above the ground is the plane?


      
      
        You are viewing the Statue of Liberty from sea level at a horizontal distance of 65 meters from the point below the torch. Standing on its pedestal above sea level, the statue subtends an angle of #math.equation(block: false, alt: "19.5 degrees")[$19.5^(∘)$], and the angle of elevation to the foot of the statue is #math.equation(block: false, alt: "32.3 degrees")[$32.3^(∘)$]. See the figure at right.


        #figure(figph[statue], alt: "statue", caption: none)


        + Find the distance #math.equation(block: false, alt: "A B")[$A B$].
+ What is the angle at #math.equation(block: false, alt: "C")[$C$]?
+ According to your measurements, how tall is the statue?


        
          + 79.64 m
+ #math.equation(block: false, alt: "35.2 degrees")[$35.2^(∘)$]
+ 46.12 m


        
      
      
        King Kong is hanging from the top of a building. From a safe distance Sherman measures that Kong subtends an angle of #math.equation(block: false, alt: "1.5 degrees")[$1.5^(∘)$], and the angle of elevation to Kong's foot is #math.equation(block: false, alt: "47.3 degrees")[$47.3^(∘)$]. See the figure at right. We know that Kong is 60 feet tall.


        #figure(figph[King Kong on building], alt: "King Kong on building", caption: none)


        + What is the angle to the top of the building (or Kong's highest point)?
+ What is the angle #math.equation(block: false, alt: "D E F")[$D E F$] at Kong's foot?
+ How long is #math.equation(block: false, alt: "D F")[$D F$], the distance from the top of the building to Sherman at #math.equation(block: false, alt: "F")[$F$]?
+ How tall is the building?


      
      Here is a surveyor's technique for making a right angle.


      + Place two stakes at points #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], 5 units apart.
+ Draw an arc of radius 4 centered at #math.equation(block: false, alt: "A")[$A$], and an arc of radius 3 centered at #math.equation(block: false, alt: "B")[$B$].
+ Place a third stake at #math.equation(block: false, alt: "C")[$C$], the intersection of the two arcs.
+ Because #math.equation(block: false, alt: "3 squared plus 4 squared equals 5 squared")[$3^(2) + 4^(2) = 5^(2)$], the angle at #math.equation(block: false, alt: "C")[$C$] is a right angle.


      #figure(figph[surveyor triangle], alt: "surveyor triangle", caption: none)


      Of course, the accuracy of the right angle constructed in this way depends on the accuracy of the measurements of length.  Problems 49 and 50 refer to the surveyor's technique.


      
        Suppose that you use the surveyor's technique to create a right angle, but all three of your distance measurements are in error. The actual sides of your triangle are #math.equation(block: false, alt: "a equals 2.9 , b equals 3.9 , c equals 5.1")[$a = 2.9 , " " b = 3.9 , " " c = 5.1$]. How far off from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] is the angle at #math.equation(block: false, alt: "C")[$C$]?


        
          #math.equation(block: true, alt: "6.1 degrees")[$6.1^(∘)$]


        
      
      
        Suppose that you use the surveyor's technique to create a right angle, but the true distance measurements are#math.equation(block: false, alt: "a equals 3.1 , b equals 4.1 , c equals 5.1")[$a = 3.1 , " " b = 4.1 , " " c = 5.1$]. How far off from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] is the angle at #math.equation(block: false, alt: "C")[$C$]?


      
      
        Triangle #math.equation(block: false, alt: "A B C")[$A B C$] has sides #math.equation(block: false, alt: "a equals 5 , b equals 8 , c equals 9")[$a = 5 , " " b = 8 , " " c = 9$]. How far from #math.equation(block: false, alt: "B")[$B$] is the foot of the altitude from point #math.equation(block: false, alt: "A")[$A$]?


        
          #figure(figph[triangle], alt: "triangle", caption: none)


          4.2


        
      
      
        Triangle #math.equation(block: false, alt: "A B C")[$A B C$] has sides #math.equation(block: false, alt: "a equals 53 , b equals 27 , c equals 71")[$a = 53 , " " b = 27 , " " c = 71$]. How far from #math.equation(block: false, alt: "C")[$C$] is the foot of the altitude from point #math.equation(block: false, alt: "A")[$A$]?


      
      
        The hour hand of Big Ben is 9 feet long, and the minute hand is 14 feet long. How far apart are the tips of the two hands at 5:00 o'clock?


        
          #figure(figph[triangle], alt: "triangle", caption: none)


          22.25 ft


        
      
      
        The largest clock in the world sits atop the Abraj Al Bait Towers in Mecca, Saudi Arabia. The clock face is 46 meters in diameter, the minute hand is 22 meters long, and the hour hand is 17 meters long. How far apart are the tips of the two hands at 8:00 o'clock?


      
      
        How far away is a star with a parallax of #math.equation(block: false, alt: "5.2 to the power ′ ′")[$5.2^(′ ′)$] when observed from opposite ends of earth's orbit?


        
          79,332.6 AU


        
      
      
        How far away is a star with a parallax of #math.equation(block: false, alt: "23 to the power ′ ′")[$23^(′ ′)$] when observed from opposite ends of earth's orbit?


      
      Problems 57–58 provide a geometric interpretation of the Law of Sines. Recall the fact from geometry that the measure of an angle inscribed in a circle is half the intercepted arc.


      
        In the figure, the circle centered at #math.equation(block: false, alt: "O")[$O$] circumscribes #math.equation(block: false, alt: "△ X Y Z")[$upright(△) X Y Z$]. Suppose that #math.equation(block: false, alt: "θ equals ∠ X")[$θ = upright(∠) X$] is an acute angle, and the side opposite #math.equation(block: false, alt: "θ")[$θ$], #math.equation(block: false, alt: "Y Z bar")[$limits(Y Z)^(―)$], has length #math.equation(block: false, alt: "s")[$s$]. Notice that #math.equation(block: false, alt: "△ O Y Z")[$upright(△) O Y Z$] is isosceles, because two of its sides are radii of the circle.


        #figure(figph[circumscribed triangle], alt: "circumscribed triangle", caption: none)


        + If #math.equation(block: false, alt: "W")[$W$] is the midpoint of #math.equation(block: false, alt: "Y Z bar")[$limits(Y Z)^(―)$], explain why #math.equation(block: false, alt: "∠ W O Y equals θ")[$upright(∠) W O Y = θ$].
+ Write #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] in terms of #math.equation(block: false, alt: "s")[$s$] and the radius #math.equation(block: false, alt: "r")[$r$]. (Hint: Use the right triangle #math.equation(block: false, alt: "W O Y")[$W O Y$] and the fact that #math.equation(block: false, alt: "W Y equals the fraction 1 over 2 s")[$W Y = display(frac(1, 2)) s$].)
+ Use your equation from (b) to write #math.equation(block: false, alt: "r")[$r$] in terms of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] and #math.equation(block: false, alt: "s")[$s$].
+ Use your equation from (c) to write the diameter #math.equation(block: false, alt: "d")[$d$] in terms of #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] and #math.equation(block: false, alt: "s")[$s$].


        
          + #math.equation(block: false, alt: "O W")[$O W$] bisects the central angle at #math.equation(block: false, alt: "O")[$O$], and the inscribed angle #math.equation(block: false, alt: "θ")[$θ$] is half the central angle at #math.equation(block: false, alt: "O")[$O$].
+ #math.equation(block: false, alt: "sin ⁡ θ equals the fraction s over 2 r")[$sin θ = display(frac(s, 2 r))$]
+ #math.equation(block: false, alt: "r equals the fraction s over 2 sin ⁡ open parenthesis θ close parenthesis")[$r = display(frac(s, 2 sin ( θ )))$]
+ #math.equation(block: false, alt: "d equals the fraction s over sin ⁡ open parenthesis θ close parenthesis")[$d = display(frac(s, sin ( θ )))$]


        
      
      
        Let #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] have lengths #math.equation(block: false, alt: "a , b")[$a , " " b$], and #math.equation(block: false, alt: "c")[$c$] opposite the angles #math.equation(block: false, alt: "A , B")[$A , " " B$], and #math.equation(block: false, alt: "C")[$C$] as usual. Use the result of Problem 57, the Law of Sines, and the fact that every triangle has an acute angle to show that the quantities #math.equation(block: false, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis")[$display(frac(a, sin ( A )))$], #math.equation(block: false, alt: "the fraction b over sin ⁡ open parenthesis B close parenthesis")[$display(frac(b, sin ( B )))$], and #math.equation(block: false, alt: "the fraction c over sin ⁡ open parenthesis C close parenthesis")[$display(frac(c, sin ( C )))$] all represent the diameter of the circumscribing circle.