solutions to exercises, section 6 - uw-madison …park/fall2014/precalculus/6.1sol.pdf ·...

Instructor’s Solutions Manual, Section 6.1 Exercise 1

Solutions to Exercises, Section 6.1

1. Find the area of a triangle that has sides of length 3 and 4, with anangle of 37◦ between those sides.

solution The area of this triangle equals 3·4·sin 37◦2 , which equals

6 sin 37◦. A calculator shows that this is approximately 3.61 (make surethat your calculator is computing in degrees, or first convert to radians,when doing this calculation).


2. Find the area of a triangle that has sides of length 4 and 5, with anangle of 41◦ between those sides.

solution This triangle has area 4·5·sin 41◦2 , which equals 10 sin 41◦. A

calculator shows that this is approximately 6.56 (make sure that yourcalculator is computing in degrees, or first convert to radians, whendoing this calculation).


3. Find the area of a triangle that has sides of length 2 and 7, with anangle of 3 radians between those sides.

solution The area of this triangle equals 2·7·sin 32 , which equals 7 sin 3.

A calculator shows that this is approximately 0.988 (make sure thatyour calculator is computing in radians, or first convert to degrees,when doing this calculation).


4. Find the area of a triangle that has sides of length 5 and 6, with anangle of 2 radians between those sides.

solution The area of this triangle equals 5·6·sin 22 , which equals

15 sin 2. A calculator shows that this is approximately 13.6395 (makesure that your calculator is computing in radians, or first convert todegrees, when doing this calculation).


For Exercises 5–12 use the following figure (which is not drawn to scale):

b

a

Θ

5. Find the value of b if a = 3, θ = 30◦, and the area of the triangle equals5.

solution Because the area of the triangle equals 5, we have

5 = ab sinθ2 = 3b sin 30◦

2 = 3b4 .

Solving the equation above for b, we get b = 203 .


6. Find the value of a if b = 5, θ = 45◦, and the area of the triangle equals8.


8 = ab sinθ2 = a5 sin 45◦

2 = 5a√

24 .

Solving the equation above for a, we get a = 16√

25 .


7. Find the value of a if b = 7, θ = π4 , and the area of the triangle equals

10.


10 = ab sinθ2 = 7a sin π

42 = 7a

2√

2.


27 .


8. Find the value of b if a = 9, θ = π3 , and the area of the triangle equals 4.


4 = ab sinθ2 = 9b sin π

32 = 9b

√3

4 .

Solving the equation above for b, we get b = 16√

327 .


9. Find the value of θ (in radians) if a = 7, b = 6, the area of the triangleequals 15, and θ < π

2 .


15 = ab sinθ2 = 7·6·sinθ

2 = 21 sinθ.

Solving the equation above for sinθ, we get sinθ = 57 . Thus

θ = sin−1 57 ≈ 0.7956.


10. Find the value of θ (in radians) if a = 5, b = 4, the area of the triangleequals 3, and θ < π

2 .



2 = 10 sinθ.

Solving the equation above for sinθ, we get sinθ = 310 . Thus

θ = sin−1 310 ≈ 0.3047.


11. Find the value of θ (in degrees) if a = 6, b = 3, the area of thetriangle equals 5, and θ > 90◦.



2 = 9 sinθ.

Solving the equation above for sinθ, we get sinθ = 59 . Thus θ equals

π − sin−1 59 radians. Converting this to degrees, we have

θ = 180◦ − (sin−1 59)

180π◦ ≈ 146.25◦.


12. Find the value of θ (in degrees) if a = 8, b = 5, and the area of thetriangle equals 12, and θ > 90◦.



2 = 20 sinθ.

Solving the equation above for sinθ, we get sinθ = 35 . Thus θ equals

π − sin−1 35 radians. Converting this to degrees, we have

θ = 180◦ − (sin−1 35)

180π◦ ≈ 143.13◦.


13. Find the area of a parallelogram that has pairs of sides of lengths 6and 9, with an angle of 81◦ between two of those sides.

solution The area of this parallelogram equals 6 · 9 · sin 81◦, whichequals 54 sin 81◦. A calculator shows that this is approximately 53.34.


14. Find the area of a parallelogram that has pairs of sides of lengths 5and 11, with an angle of 28◦ between two of those sides.

solution The area of this parallelogram equals 5 · 11 · sin 28◦, whichequals 55 sin 28◦. A calculator shows that this is approximately 25.82.


15. Find the area of a parallelogram that has pairs of sides of lengths 4 and10, with an angle of π

6 radians between two of those sides.

solution The area of this parallelogram equals 4 · 10 · sin π6 , which

equals 20.


16. Find the area of a parallelogram that has pairs of sides of lengths 3 and12, with an angle of π

3 radians between two of those sides.

solution The area of this parallelogram equals 3 · 12 · sin π3 , which

equals 18√

3.


For Exercises 17–24, use the following figure (which is not drawn to scaleexcept that u is indeed meant to be an acute angle and ν is indeedmeant to be an obtuse angle):

b

b

aa

u Ν

17. Find the value of b if a = 4, ν = 135◦, and the area of the parallelogramequals 7.

solution Because the area of the parallelogram equals 7, we have

7 = ab sinν = 4b sin 135◦ = 2√

2b.


2= 7

√2

4 .


18. Find the value of a if b = 6, ν = 120◦, and the area of the parallelogramequals 11.


11 = ab sinν = a6 sin 120◦ = 3√

3a.


3= 11

√3

9 .


19. Find the value of a if b = 10, u = π3 , and the area of the parallelogram

equals 7.


7 = ab sinu = 10a sin π3 = 5a

√3.


3= 7

√3

15 .


20. Find the value of b if a = 5, u = π4 , and the area of the parallelogram

equals 9.


9 = ab sinu = 5b sin π4 = 5b

√2

2 .


2= 9

√2

5 .


21. Find the value of u (in radians) if a = 3, b = 4, and the area of theparallelogram equals 10.


10 = ab sinu = 3 · 4 · sinu = 12 sinu.

Solving the equation above for sinu, we get sinu = 56 . Thus

u = sin−1 56 ≈ 0.9851.


22. Find the value of u (in radians) if a = 4, b = 6, and the area of theparallelogram equals 19.


19 = ab sinu = 4 · 6 · sinu = 24 sinu.

Solving the equation above for sinu, we get sinu = 1924 . Thus

u = sin−1 1924 ≈ 0.9135.


23. Find the value of ν (in degrees) if a = 6, b = 7, and the area of theparallelogram equals 31.


31 = ab sinν = 6 · 7 · sinν = 42 sinν.

Solving the equation above for sinν , we get sinν = 3142 . Because ν is an

obtuse angle, we thus have ν = π − sin−1 3142 radians. Converting this to

degrees, we have ν = 180◦ − (sin−1 3142)

180π◦ ≈ 132.43◦.


24. Find the value of ν (in degrees) if a = 8, b = 5, and the area of theparallelogram equals 12.


12 = ab sinν = 8 · 5 · sinν = 40 sinν.

Solving the equation above for sinν , we get sinν = 310 . Because ν is an

obtuse angle, we thus have ν = π − sin−1 310 radians. Converting this to

degrees, we have ν = 180◦ − (sin−1 310)

180π◦ ≈ 162.54◦.


25. What is the largest possible area for a triangle that has one side oflength 4 and one side of length 7?

solution In a triangle that has one side of length 4 and one side oflength 7, let θ denote the angle between those two sides. Thus the areaof the triangle will equal

14 sinθ.

We need to choose θ to make this area as large as possible. The largestpossible value of sinθ is 1, which occurs when θ = π

2 (or θ = 90◦ if weare working in degrees). Thus we choose θ = π

2 , which gives us a righttriangle with sides of length 4 and 7 around the right angle.

4

7

This right triangle has area 14, which is thelargest area of any triangle with sides oflength 4 and 7.


26. What is the largest possible area for a parallelogram that has pairs ofsides with lengths 5 and 9?

solution In a parallelogram that has pairs of sides with lengths 5 and9, let θ denote an angle between two adjacent sides. Thus the area ofthe parallelogram will equal

45 sinθ.

We need to choose θ to make this area as large as possible. The largestpossible value of sinθ is 1, which occurs when θ = π

2 (or θ = 90◦ if weare working in degrees). Thus we choose θ = π

2 , which gives us arectangle with sides of length 5 and 9.

5

9

This rectangle has area 45, which is thelargest area of any parallelogram with sides oflength 5 and 9.


27. Sketch the regular hexagon whose vertices are six equally spaced pointson the unit circle, with one of the vertices at the point (1,0).

solution

1


28. Sketch the regular dodecagon whose vertices are twelve equally spacedpoints on the unit circle, with one of the vertices at the point (1,0).[A dodecagon is a twelve-sided polygon.]

solution

1


29. Find the coordinates of all six vertices of the regular hexagon whosevertices are six equally spaced points on the unit circle, with (1,0) asone of the vertices. List the vertices in counterclockwise order startingat (1,0).

solution The coordinates of the six vertices, listed incounterclockwise order starting at (1,0), are (cos 2πm

6 , sin 2πm6 ), with

m going from 0 to 5. Evaluating the trigonometric functions, we get thefollowing list of coordinates of vertices: (1,0), (1

2 ,√

32 ), (−1

2 ,√

32 ),

(−1,0), (−12 ,−

√3

2 ), (12 ,−

√3

2 ).


30. Find the coordinates of all twelve vertices of the dodecagondodecagonwhose vertices are twelve equally spaced points on the unit circle, with(1,0) as one of the vertices. List the vertices in counterclockwise orderstarting at (1,0).

solution The coordinates of the twelve vertices, listed incounterclockwise order starting at (1,0), are (cos 2πm

12 , sin 2πm12 ), with

m going from 0 to 11. Evaluating the trigonometric functions, we getthe following list of coordinates of vertices: (1,0), (

√3

2 ,12), (

12 ,√

32 ),

(0,1), (−12 ,√

32 ), (−

√3

2 ,12), (−1,0), (−

√3

2 ,−12), (−1

2 ,−√

32 ), (0,−1),

(12 ,−

√3

2 ), (√

32 ,−1

2).


31. Find the area of a regular hexagon whose vertices are six equally spacedpoints on the unit circle.

solution Decompose the hexagon into triangles by drawing linesegments from the center of the circle (the origin) to the vertices. Eachtriangle has two sides that are radii of the unit circle; thus those twosides of the triangle each have length 1. The angle between those tworadii is 2π

6 radians (because one rotation around the entire circle is anangle of 2π radians, and each of the six triangles has an angle thattakes up one-sixth of the total). Now 2π

6 radians equals π3 radians (or

60◦). Thus each of the six triangles has area

12 · 1 · 1 · sin π

3 ,

which equals√

34 . Thus the sum of the areas of the six triangles equals

6 ·√

34 , which equals 3

√3

2 . In other words, the hexagon has area 3√

32 .


32. Find the area of a regular dodecagon whose vertices are twelve equallyspaced points on the unit circle.

solution Decompose the dodecagon into triangles by drawing linesegments from the center of the circle (the origin) to the vertices. Eachtriangle has two sides that are radii of the unit circle; thus those twosides of the triangle each have length 1. The angle between those tworadii is 2π

12 radians (because one rotation around the entire circle is anangle of 2π radians, and each of the twelve triangles has an angle thattakes up one-twelfth of the total). Now 2π

12 radians equals π6 radians (or

30◦). Thus each of the twelve triangles has area

12 · 1 · 1 · sin π

6 ,

which equals 14 . Thus the sum of the areas of the twelve triangles

equals 12 · 14 , which equals 3. In other words, the dodecagon has area 3.


33. Find the perimeter of a regular hexagon whose vertices are six equallyspaced points on the unit circle.

solution If we assume that one of the vertices of the hexagon is thepoint (1,0), then the next vertex in the counterclockwise direction isthe point (1

2 ,√

32 ). Thus the length of each side of the hexagon equals

the distance between (1,0) and (12 ,√

32 ), which equals

√(1− 1

2

)2 + (√32

)2,

which equals 1. Thus the perimeter of the hexagon equals 6 · 1, whichequals 6.


34. Find the perimeter of a regular dodecagondodecagon whose verticesare twelve equally spaced points on the unit circle.

solution If we assume that one of the vertices of the dodecagon isthe point (1,0), then the next vertex in the counterclockwise directionis the point (

√3

2 ,12). Thus the length of each side of the dodecagon

equals the distance between (1,0) and (√

32 ,

12), which equals

√(1−

√3

2

)2 + (12

)2,

which equals√

2−√3. Thus the perimeter of the dodecagon equals12√

2−√3.


35. Find the area of a regular hexagon with sides of length s.

solution There is a constant c such that a regular hexagon with sidesof length s has area cs2. From Exercises 31 and 33, we know that thearea equals 3

√3

2 if s = 1. Thus

3√

32 = c · 12 = c.

Thus a regular hexagon with sides of length s has area 3√

32 s2.


36. Find the area of a regular dodecagon with sides of length s.

solution There is a constant c such that a regular dodecagon withsides of length s has area cs2. From Exercises 32 and 34, we know thatthe area equals 3 if s =

√2−√3. Thus

3 = c(√

2−√3)2 = c(2−√3).

Solving this equation for c, we have

c = 3

2−√3= 3

2−√3· 2+√3

2+√3= 6+ 3

√3.

Thus a regular dodecagon with sides of length s has area (6+ 3√

3)s2.


37. Find the area of a regular 13-sided polygon whose vertices are 13equally spaced points on a circle of radius 4.

solution Decompose the 13-sided polygon into triangles by drawingline segments from the center of the circle to the vertices. Each trianglehas two sides that are radii of the circle with radius 4; thus those twosides of the triangle each have length 4. The angle between those tworadii is 2π

13 radians (because one rotation around the entire circle is anangle of 2π radians, and each of the 13 triangles has an angle that takesup one-thirteenth of the total). Thus each of the 13 triangles has area

12 · 4 · 4 · sin 2π

13 ,

which equals 8 sin 2π13 . The area of the 13-sided polygon is the sum of

the areas of the 13 triangles, which equals 13 · 8 sin 2π13 , which is

approximately 48.3.


38. The face of a Canadian one-dollar coin is a regular 11-sided polygon(see the picture just before the start of these exercises). The distancefrom the center of this polygon to one of the vertices is 1.325centimeters. Find the area of the face of this coin.

solution Think of the face of the coin as being inscribed in a circlewith radius 1.325 centimeters. Decompose the 11-sided polygon intotriangles by drawing line segments from the center of the circle to thevertices. Each triangle has two sides that are radii of the circle withradius 1.325 centimeters; thus those two sides of the triangle each havelength 1.325 centimeters. The angle between those two radii is 2π

11radians (because one rotation around the entire circle is an angle of 2πradians, and each of the 11 triangles has an angle that takes upone-eleventh of the total). Thus each of the 11 triangles has area

12 · 1.325 · 1.325 · sin 2π

11 ,

square centimeters. The area of the 11-sided polygon is the sum of theareas of the 11 triangles, which equals 11 · 1

2 · 1.325 · 1.325 · sin 2π11

square centimeters, which is approximately 5.22 square centimeters.

Instructor’s Solutions Manual, Section 6.1 Problem 39

Solutions to Problems, Section 6.1

39. What is the area of a triangle whose sides all have length r?

solution A triangle all of whose sides have length r is an equilateraltriangle all of whose angles are π

3 radians (or 60◦). Thus the area of

such a triangle is 12r

2 sin π3 , which equals

√3r2

4 .


40. Explain why there does not exist a triangle with area 15 having one sideof length 4 and one side of length 7.

solution Consider a triangle with one side of length 4, one side oflength 7, and an angle θ between these two sides. The area of thistriangle equals 1

2 · 4 · 7 sinθ, which equals 14 sinθ. Because sinθ ≤ 1,the area of this triangle is less than or equal to 14. Thus this trianglecannot have area 15.


41. Show that if a triangle has area R, sides of length A, B, and C , andangles a, b, and c, then

R3 = 18A

2B2C2(sina)(sinb)(sin c).

[Hint: Write three formulas for the area R, and then multiply theseformulas together.]

solution Consider a triangle with sides of length A, B, and C . Let abe the angle between the sides of length B and C , let b be the anglebetween the sides of length A and C , and let c be the angle between thesides of length A and B. Let R be the area of this triangle. Then we havethe formulas

R = 12BC sina

R = 12AC sinb

R = 12AB sin c.

Multiplying these three formulas together gives

R3 = 18A

2B2C2(sina)(sinb)(sin c).


42. Find numbers b and c such that an isosceles triangle with sides oflength b, b, and c has perimeter and area that are both integers.

solution Consider an isosceles triangle with sides of length b, b, andc. The perimeter of this triangle is 2b + c.

Let h denote the height of this triangle as shown in the figure below.The area of this triangle is ch

2 . As can be seen in the figure below, wehave a right triangle with sides of length h, c2 , and b. To makeeverything in sight an integer, we will make this be the right trianglewith sides of length 3, 4, and 5. In other words, we choose b = 5 andc = 8 (which makes c

2 = 4 and, by the Pythagorean Theorem, makesh = 4).

b b

c�2

h

With b = 5 and c = 8, this triangle has perimeter 18 and area 20.

Of course there are also other correct solutions.


43. Explain why the solution to Exercise 32 is somewhat close to π .

solution The area inside a circle of radius 1 is π , which isapproximately 3.14. The regular dodecagon in Exercise 32 fills up mostof the area inside the unit circle, so its area should be just somewhatless that the area inside the circle. Indeed, the dodecagon in Exercise 32has area 3, so it misses about 0.14 of the area inside the circle.


44. Use a calculator to evaluate numerically the exact solution youobtained to Exercise 34. Then explain why this number is somewhatclose to 2π .

solution In Exercise 34 we found that a dodecagon whose verticesare equally spaced points on the unit circle has perimeter 12

√2−√3. A

calculator shows that this number is approximately 6.21.

The perimeter of this dodecagon should be just somewhat less than thecircumference of the unit circle, as can be seen in the figure that is thesolution to Exercise 28. The unit circle has circumference 2π , which isapproximately 6.28. Thus the perimeter of the dodecagon(approximately 6.21) is somewhat close to the circumference of thecircle (approximately 6.28) as we indeed expect from the figure.


45. Explain why a regular polygon with n sides whose vertices are nequally spaced points on the unit circle has area n

2 sin 2πn .

solution Consider a regular polygon with n sides whose vertices aren equally spaced points on the unit circle. Draw a radius from theorigin to each vertex, partitioning the polygon into n isoscelestriangles. Each of these n isosceles triangles has two sides of length 1(the radii), with angle 2π

n between these two sides (because the entirecircle, which has angle 2π , has been partitioned into n equal angles).Thus the area of each of the n isosceles triangles is 1

2 · 1 · 1 · sin 2πn ,

which equals 12 sin 2π

n .

Because the regular polygon with n sides is composed of n of thesetriangles, the area of the polygon equals n

2 sin 2πn .


46. Explain why the result stated in the previous problem implies that

sin 2πn ≈ 2π

n

for large positive integers n.

solution Suppose n is a large positive integer. Then a regularpolygon with n sides whose vertices are n equally spaced points on theunit circle fills up almost all the area inside the unit circle. Thus thearea of this polygon, which from the previous problem equals n

2 sin 2πn ,

should be approximately the same as the area inside the unit circle,which equals π . In other words, we have

n2 sin 2π

n ≈ π.

Multiplying both sides by 2n gives the approximation

sin 2πn ≈ 2π

n .


47. Choose three large values of n, and use a calculator to verify thatsin 2π

n ≈ 2πn for each of those three large values of n.

solution The table below gives the values of sin 2πn and 2π

n to sixsignificant digits for n = 1000, n = 10000, and n = 100000:

n sin 2πn

2πn

1000 0.00628314 0.00628319

10000 0.000628318 0.000628319

100000 0.0000628319 0.0000628319

As can be seen from the table above, we have sin 2πn ≈ 2π

n for thesethree large values of n, with the approximation more accurate forlarger values of n.


48. Show that each edge of a regular polygon with n sides whose verticesare n equally spaced points on the unit circle has length√

2− 2 cos 2πn .

solution Consider a regular polygon with n sides whose vertices aren equally spaced points on the unit circle, with one of the vertices atthe point (1,0). The next vertex in the counterclockwise direction hascoordinates

(cos 2π

n , sin 2πn), because the radius ending at that vertex

has angle 2πn with the positive horizontal axis.

Thus the length of this edge of the polygon (and hence of each edge ofthe polygon) is the distance between the points (1,0) and(cos 2π

n , sin 2πn). We can compute that distance using the usual formula

for the distance between two points. Thus the length of each edge ofthis polygon is given by the following formula:√(

1− cos 2πn)2 + sin2 2π

n =√

1− 2 cos 2πn + cos2 2π

n + sin2 2πn

=√

2− 2 cos 2πn


49. Explain why a regular polygon with n sides, each with length s, has area

n sin 2πn

4(1− cos 2πn )s2.

solution A regular polygon with n sides, with the vertices equallyspaced on the unit circle, has area n

2 sin 2πn (from Problem 45) and has

sides of length√

2− 2 cos 2πn (from Problem 48).

Thus to get a regular polygon with sides having length s, wehorizontally and vertically stretch this polygon with vertices on the unitcircle by a factor of

s√2− 2 cos 2π

n

.

By the Area Stretch Theorem (Section 4.2), this changes the area by afactor of (

s√2− 2 cos 2π

n

)2

.

Thus a regular polygon with n sides, each with length s, has area

n2 (sin 2π

n )(

s√2− 2 cos 2π

n

)2

,

which equals


n sin 2πn

4(1− cos 2πn )s2.


50. Verify that for n = 4, the formula given by the previous problemreduces to the usual formula for the area of a square.

solution If n = 4, then

n sin 2πn

4(1− cos 2πn )s2 = 4 sin π

2

4(1− cos π2 )s2

= s2,

which is the usual formula for the area of a square with sides oflength s.


51. Explain why a regular polygon with n sides whose vertices are nequally spaced points on the unit circle has perimeter

n√

2− 2 cos 2πn .

solution By Problem 48, we know that each side of a regular polygonwith n sides whose vertices lie on the unit circle has length√

2− 2 cos 2πn . The perimeter of the polygon is n times the length of

each side. Thus the perimeter of this polygon is

n√

2− 2 cos 2πn .


52. Explain why the result stated in the previous problem implies that

n√

2− 2 cos 2πn ≈ 2π

for large positive integers n.

solution If n is a large positive integer, then the perimeter of regularpolygon with n sides whose vertices lie on the unit circle isapproximately equal to the circumference of the unit circle, whichequals 2π . In other words, if n is a large positive integer then

n√

2− 2 cos 2πn ≈ 2π.


53. Choose three large values of n, and use a calculator to verify that

n√

2− 2 cos 2πn ≈ 2π for each of those three large values of n.

solution The table below gives the values of n√

2− 2 cos 2πn and 2π

to six significant digits for n = 100, n = 1000, and n = 10000:

n n√

2− 2 cos 2πn 2π

100 6.28215 6.28319

1000 6.28317 6.28319

10000 6.28319 6.28319

As can be seen from the table above, we have n√

2− 2 cos 2πn ≈ 2π for

these three large values of n, with the approximation more accurate forlarger values of n.


54. Show thatcos 2π

n ≈ 1− 2π2

n2

if n is a large positive integer.

solution Suppose n is a large positive integer. Then we know fromProblem 52 that

n√

2− 2 cos 2πn ≈ 2π.

Square both sides of the approximation above and then divide by n2 toget

2− 2 cos 2πn ≈ 4π2

n2 .

Now divide both sides by 2 and then solve for cos 2πn , getting

cos 2πn ≈ 1− 2π2

n2 .

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