Solutions to Exercises, Section 6.1


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1 Instructor s Solutions Manual, Section 6.1 Exercise 1 Solutions to Exercises, Section Find the area of a triangle that has sides of length 3 and 4, with an angle of 37 between those sides. 3 4 sin 37 solution The area of this triangle equals, which equals 6 sin 37. A calculator shows that this is approximately 3.61 (make sure that your calculator is computing in degrees, or first convert to radians, when doing this calculation).
2 Instructor s Solutions Manual, Section 6.1 Exercise. Find the area of a triangle that has sides of length 4 and 5, with an angle of 41 between those sides. 4 5 sin 41 solution This triangle has area, which equals 10 sin 41.A calculator shows that this is approximately 6.56 (make sure that your calculator is computing in degrees, or first convert to radians, when doing this calculation).
3 Instructor s Solutions Manual, Section 6.1 Exercise 3 3. Find the area of a triangle that has sides of length and 7, with an angle of 3 radians between those sides. solution The area of this triangle equals 7 sin 3, which equals 7 sin 3. A calculator shows that this is approximately (make sure that your calculator is computing in radians, or first convert to degrees, when doing this calculation).
4 Instructor s Solutions Manual, Section 6.1 Exercise 4 4. Find the area of a triangle that has sides of length 5 and 6, with an angle of radians between those sides. solution The area of this triangle equals 5 6 sin, which equals 15 sin. A calculator shows that this is approximately (make sure that your calculator is computing in radians, or first convert to degrees, when doing this calculation).
5 Instructor s Solutions Manual, Section 6.1 Exercise 5 For Exercises 5 1 use the following figure (which is not drawn to scale): a Θ b 5. Find the value of b if a = 3, θ = 30, and the area of the triangle equals 5. solution Because the area of the triangle equals 5, we have 5 = ab sin θ 3b sin 30 = = 3b 4. Solving the equation above for b, wegetb = 0 3.
6 Instructor s Solutions Manual, Section 6.1 Exercise 6 6. Find the value of a if b = 5, θ = 45, and the area of the triangle equals 8. solution Because the area of the triangle equals 8, we have 8 = ab sin θ = a5sin45 = 5a 4. Solving the equation above for a, wegeta = 16 5.
7 Instructor s Solutions Manual, Section 6.1 Exercise 7 7. Find the value of a if b = 7, θ = π 4, and the area of the triangle equals 10. solution Because the area of the triangle equals 10, we have 10 = ab sin θ = 7a sin π 4 = 7a. Solving the equation above for a, wegeta = 0 7.
8 Instructor s Solutions Manual, Section 6.1 Exercise 8 8. Find the value of b if a = 9, θ = π 3, and the area of the triangle equals 4. solution Because the area of the triangle equals 4, we have 4 = ab sin θ = 9b sin π 3 = 9b 3 4. Solving the equation above for b, wegetb =
9 Instructor s Solutions Manual, Section 6.1 Exercise 9 9. Find the value of θ (in radians) if a = 7, b = 6, the area of the triangle equals 15, and θ< π. solution Because the area of the triangle equals 15, we have 15 = ab sin θ = 7 6 sin θ = 1 sin θ. Solving the equation above for sin θ, we get sin θ = 5 7. Thus θ = sin
10 Instructor s Solutions Manual, Section 6.1 Exercise Find the value of θ (in radians) if a = 5, b = 4, the area of the triangle equals 3, and θ< π. solution Because the area of the triangle equals 3, we have 3 = ab sin θ = 5 4 sin θ = 10 sin θ. Solving the equation above for sin θ, we get sin θ = Thus θ = sin
11 Instructor s Solutions Manual, Section 6.1 Exercise Find the value of θ (in degrees) if a = 6, b = 3, the area of the triangle equals 5, and θ>90. solution Because the area of the triangle equals 5, we have 5 = ab sin θ = 6 3 sin θ = 9 sin θ. Solving the equation above for sin θ, we get sin θ = 5 9. Thus θ equals π sin radians. Converting this to degrees, we have θ = 180 (sin ) 180 π
12 Instructor s Solutions Manual, Section 6.1 Exercise 1 1. Find the value of θ (in degrees) if a = 8, b = 5, and the area of the triangle equals 1, and θ>90. solution Because the area of the triangle equals 1, we have 1 = ab sin θ = 8 5 sin θ = 0 sin θ. Solving the equation above for sin θ, we get sin θ = 3 5. Thus θ equals π sin radians. Converting this to degrees, we have θ = 180 (sin ) 180 π
13 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a parallelogram that has pairs of sides of lengths 6 and 9, with an angle of 81 between two of those sides. solution The area of this parallelogram equals 6 9 sin 81, which equals 54 sin 81. A calculator shows that this is approximately
14 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a parallelogram that has pairs of sides of lengths 5 and 11, with an angle of 8 between two of those sides. solution The area of this parallelogram equals 5 11 sin 8, which equals 55 sin 8. A calculator shows that this is approximately 5.8.
15 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a parallelogram that has pairs of sides of lengths 4 and 10, with an angle of π 6 radians between two of those sides. solution equals 0. The area of this parallelogram equals 4 10 sin π 6, which
16 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a parallelogram that has pairs of sides of lengths 3 and 1, with an angle of π 3 radians between two of those sides. solution The area of this parallelogram equals 3 1 sin π 3, which equals 18 3.
17 Instructor s Solutions Manual, Section 6.1 Exercise 17 For Exercises 17 4, use the following figure (which is not drawn to scale except that u is indeed meant to be an acute angle and ν is indeed meant to be an obtuse angle): b a a u b Ν 17. Find the value of b if a = 4, ν = 135, and the area of the parallelogram equals 7. solution Because the area of the parallelogram equals 7, we have 7 = ab sin ν = 4b sin 135 = b. Solving the equation above for b, wegetb = 7 = 7 4.
18 Instructor s Solutions Manual, Section 6.1 Exercise Find the value of a if b = 6, ν = 10, and the area of the parallelogram equals 11. solution Because the area of the parallelogram equals 11, we have 11 = ab sin ν = a6 sin 10 = 3 3a. Solving the equation above for a, wegeta = =
19 Instructor s Solutions Manual, Section 6.1 Exercise Find the value of a if b = 10, u = π 3, and the area of the parallelogram equals 7. solution Because the area of the parallelogram equals 7, we have 7 = ab sin u = 10a sin π 3 = 5a 3. Solving the equation above for a, wegeta = =
20 Instructor s Solutions Manual, Section 6.1 Exercise 0 0. Find the value of b if a = 5, u = π 4, and the area of the parallelogram equals 9. solution Because the area of the parallelogram equals 9, we have 9 = ab sin u = 5b sin π 4 = 5b. Solving the equation above for b, wegetb = 18 5 = 9 5.
21 Instructor s Solutions Manual, Section 6.1 Exercise 1 1. Find the value of u (in radians) if a = 3, b = 4, and the area of the parallelogram equals 10. solution Because the area of the parallelogram equals 10, we have 10 = ab sin u = 3 4 sin u = 1 sin u. Solving the equation above for sin u, we get sin u = 5 6. Thus u = sin
22 Instructor s Solutions Manual, Section 6.1 Exercise. Find the value of u (in radians) if a = 4, b = 6, and the area of the parallelogram equals 19. solution Because the area of the parallelogram equals 19, we have 19 = ab sin u = 4 6 sin u = 4 sin u. Solving the equation above for sin u, we get sin u = Thus u = sin
23 Instructor s Solutions Manual, Section 6.1 Exercise 3 3. Find the value of ν (in degrees) if a = 6, b = 7, and the area of the parallelogram equals 31. solution Because the area of the parallelogram equals 31, we have 31 = ab sin ν = 6 7 sin ν = 4 sin ν. Solving the equation above for sin ν, we get sin ν = Because ν is an 1 31 obtuse angle, we thus have ν = π sin 4 radians. Converting this to degrees, we have ν = (sin 4 ) 180 π
24 Instructor s Solutions Manual, Section 6.1 Exercise 4 4. Find the value of ν (in degrees) if a = 8, b = 5, and the area of the parallelogram equals 1. solution Because the area of the parallelogram equals 1, we have 1 = ab sin ν = 8 5 sin ν = 40 sin ν. Solving the equation above for sin ν, we get sin ν = Because ν is an obtuse angle, we thus have ν = π sin radians. Converting this to degrees, we have ν = 180 (sin ) 180 π
25 Instructor s Solutions Manual, Section 6.1 Exercise 5 5. What is the largest possible area for a triangle that has one side of length 4 and one side of length 7? solution In a triangle that has one side of length 4 and one side of length 7, let θ denote the angle between those two sides. Thus the area of the triangle will equal 14 sin θ. We need to choose θ to make this area as large as possible. The largest possible value of sin θ is 1, which occurs when θ = π (or θ = 90 if we are working in degrees). Thus we choose θ = π, which gives us a right triangle with sides of length 4 and 7 around the right angle. 4 7 This right triangle has area 14, which is the largest area of any triangle with sides of length 4 and 7.
26 Instructor s Solutions Manual, Section 6.1 Exercise 6 6. What is the largest possible area for a parallelogram that has pairs of sides with lengths 5 and 9? solution In a parallelogram that has pairs of sides with lengths 5 and 9, let θ denote an angle between two adjacent sides. Thus the area of the parallelogram will equal 45 sin θ. We need to choose θ to make this area as large as possible. The largest possible value of sin θ is 1, which occurs when θ = π (or θ = 90 if we are working in degrees). Thus we choose θ = π, which gives us a rectangle with sides of length 5 and This rectangle has area 45, which is the largest area of any parallelogram with sides of length 5 and 9.
27 Instructor s Solutions Manual, Section 6.1 Exercise 7 7. Sketch the regular hexagon whose vertices are six equally spaced points on the unit circle, with one of the vertices at the point (1, 0). solution 1
28 Instructor s Solutions Manual, Section 6.1 Exercise 8 8. Sketch the regular dodecagon whose vertices are twelve equally spaced points on the unit circle, with one of the vertices at the point (1, 0). [A dodecagon is a twelvesided polygon.] solution 1
29 Instructor s Solutions Manual, Section 6.1 Exercise 9 9. Find the coordinates of all six vertices of the regular hexagon whose vertices are six equally spaced points on the unit circle, with (1, 0) as one of the vertices. List the vertices in counterclockwise order starting at (1, 0). solution The coordinates of the six vertices, listed in counterclockwise order starting at (1, 0), are (cos πm πm 6, sin 6 ), with m going from 0 to 5. Evaluating the trigonometric functions, we get the following list of coordinates of vertices: (1, 0), ( 1, 3 ), ( 1, 3 ), ( 1, 0), ( 1, 3 ), ( 1, 3 ).
30 Instructor s Solutions Manual, Section 6.1 Exercise Find the coordinates of all twelve vertices of the dodecagondodecagon whose vertices are twelve equally spaced points on the unit circle, with (1, 0) as one of the vertices. List the vertices in counterclockwise order starting at (1, 0). solution The coordinates of the twelve vertices, listed in counterclockwise order starting at (1, 0), are (cos πm πm 1, sin 1 ), with m going from 0 to 11. Evaluating the trigonometric functions, we get 3 the following list of coordinates of vertices: (1, 0), (, 1 ), ( 1, 3 ), (0, 1), ( 1, 3 ), ( 3, 1 ), ( 1, 0), ( 3, 1 ), ( 1, 3 ), (0, 1), ( 1, 3 ), ( 3, 1 ).
31 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a regular hexagon whose vertices are six equally spaced points on the unit circle. solution Decompose the hexagon into triangles by drawing line segments from the center of the circle (the origin) to the vertices. Each triangle has two sides that are radii of the unit circle; thus those two sides of the triangle each have length 1. The angle between those two radii is π 6 radians (because one rotation around the entire circle is an angle of π radians, and each of the six triangles has an angle that takes up onesixth of the total). Now π 6 radians equals π 3 radians (or 60 ). Thus each of the six triangles has area sin π 3, which equals. Thus the sum of the areas of the six triangles equals 3 6, which equals 3 3. In other words, the hexagon has area
32 Instructor s Solutions Manual, Section 6.1 Exercise 3 3. Find the area of a regular dodecagon whose vertices are twelve equally spaced points on the unit circle. solution Decompose the dodecagon into triangles by drawing line segments from the center of the circle (the origin) to the vertices. Each triangle has two sides that are radii of the unit circle; thus those two sides of the triangle each have length 1. The angle between those two radii is π 1 radians (because one rotation around the entire circle is an angle of π radians, and each of the twelve triangles has an angle that takes up onetwelfth of the total). Now π 1 radians equals π 6 radians (or 30 ). Thus each of the twelve triangles has area sin π 6, which equals 1 4. Thus the sum of the areas of the twelve triangles equals 1 1 4, which equals 3. In other words, the dodecagon has area 3.
33 Instructor s Solutions Manual, Section 6.1 Exercise Find the perimeter of a regular hexagon whose vertices are six equally spaced points on the unit circle. solution If we assume that one of the vertices of the hexagon is the point (1, 0), then the next vertex in the counterclockwise direction is the point ( 1, 3 ). Thus the length of each side of the hexagon equals the distance between (1, 0) and ( 1, 3 ), which equals (1 1 ) ( + 3 ), which equals 1. Thus the perimeter of the hexagon equals 6 1, which equals 6.
34 Instructor s Solutions Manual, Section 6.1 Exercise Find the perimeter of a regular dodecagondodecagon whose vertices are twelve equally spaced points on the unit circle. solution If we assume that one of the vertices of the dodecagon is the point (1, 0), then the next vertex in the counterclockwise direction 3 is the point (, 1 ). Thus the length of each side of the dodecagon 3 equals the distance between (1, 0) and (, 1 ), which equals (1 3 ) ( + 1 ), which equals 3. Thus the perimeter of the dodecagon equals 1 3.
35 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a regular hexagon with sides of length s. solution There is a constant c such that a regular hexagon with sides of length s has area cs. From Exercises 31 and 33, we know that the area equals 3 3 if s = 1. Thus 3 3 = c 1 = c. Thus a regular hexagon with sides of length s has area 3 3 s.
36 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a regular dodecagon with sides of length s. solution There is a constant c such that a regular dodecagon with sides of length s has area cs. From Exercises 3 and 34, we know that the area equals 3 if s = 3. Thus Solving this equation for c, we have c = 3 = c( 3) = c( 3). 3 3 = = Thus a regular dodecagon with sides of length s has area ( )s.
37 Instructor s Solutions Manual, Section 6.1 Exercise Find the area of a regular 13sided polygon whose vertices are 13 equally spaced points on a circle of radius 4. solution Decompose the 13sided polygon into triangles by drawing line segments from the center of the circle to the vertices. Each triangle has two sides that are radii of the circle with radius 4; thus those two sides of the triangle each have length 4. The angle between those two radii is π 13 radians (because one rotation around the entire circle is an angle of π radians, and each of the 13 triangles has an angle that takes up onethirteenth of the total). Thus each of the 13 triangles has area 1 π 4 4 sin 13, which equals 8 sin π 13. The area of the 13sided polygon is the sum of the areas of the 13 triangles, which equals 13 8 sin π 13, which is approximately 48.3.
38 Instructor s Solutions Manual, Section 6.1 Exercise The face of a Canadian onedollar coin is a regular 11sided polygon (see the picture just before the start of these exercises). The distance from the center of this polygon to one of the vertices is 1.35 centimeters. Find the area of the face of this coin. solution Think of the face of the coin as being inscribed in a circle with radius 1.35 centimeters. Decompose the 11sided polygon into triangles by drawing line segments from the center of the circle to the vertices. Each triangle has two sides that are radii of the circle with radius 1.35 centimeters; thus those two sides of the triangle each have length 1.35 centimeters. The angle between those two radii is π 11 radians (because one rotation around the entire circle is an angle of π radians, and each of the 11 triangles has an angle that takes up oneeleventh of the total). Thus each of the 11 triangles has area 1 π sin 11, square centimeters. The area of the 11sided polygon is the sum of the areas of the 11 triangles, which equals 11 1 π sin 11 square centimeters, which is approximately 5. square centimeters.
39 Instructor s Solutions Manual, Section 6.1 Problem 39 Solutions to Problems, Section 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 equilateral triangle all of whose angles are π 3 radians (or 60 ). Thus the area of such a triangle is 1 r sin π 3, which equals 3r 4.
40 Instructor s Solutions Manual, Section 6.1 Problem Explain why there does not exist a triangle with area 15 having one side of length 4 and one side of length 7. solution Consider a triangle with one side of length 4, one side of length 7, and an angle θ between these two sides. The area of this triangle equals sin θ, which equals 14 sin θ. Because sin θ 1, the area of this triangle is less than or equal to 14. Thus this triangle cannot have area 15.
41 Instructor s Solutions Manual, Section 6.1 Problem Show that if a triangle has area R, sides of length A, B, and C, and angles a, b, and c, then R 3 = 1 8 A B C (sin a)(sin b)(sin c). [Hint: Write three formulas for the area R, and then multiply these formulas together.] solution Consider a triangle with sides of length A, B, and C. Let a be the angle between the sides of length B and C, let b be the angle between the sides of length A and C, and let c be the angle between the sides of length A and B. Let R be the area of this triangle. Then we have the formulas R = 1 BC sin a R = 1 AC sin b R = 1 AB sin c. Multiplying these three formulas together gives R 3 = 1 8 A B C (sin a)(sin b)(sin c).
42 Instructor s Solutions Manual, Section 6.1 Problem 4 4. Find numbers b and c such that an isosceles triangle with sides of length b, b, and c has perimeter and area that are both integers. solution Consider an isosceles triangle with sides of length b, b, and c. The perimeter of this triangle is b + c. Let h denote the height of this triangle as shown in the figure below. The area of this triangle is ch. As can be seen in the figure below, we have a right triangle with sides of length h, c, and b. To make everything in sight an integer, we will make this be the right triangle with sides of length 3, 4, and 5. In other words, we choose b = 5 and c = 8 (which makes c = 4 and, by the Pythagorean Theorem, makes h = 4). b h b c With b = 5 and c = 8, this triangle has perimeter 18 and area 0. Of course there are also other correct solutions.
43 Instructor s Solutions Manual, Section 6.1 Problem Explain why the solution to Exercise 3 is somewhat close to π. solution The area inside a circle of radius 1 is π, which is approximately The regular dodecagon in Exercise 3 fills up most of the area inside the unit circle, so its area should be just somewhat less that the area inside the circle. Indeed, the dodecagon in Exercise 3 has area 3, so it misses about 0.14 of the area inside the circle.
44 Instructor s Solutions Manual, Section 6.1 Problem Use a calculator to evaluate numerically the exact solution you obtained to Exercise 34. Then explain why this number is somewhat close to π. solution In Exercise 34 we found that a dodecagon whose vertices are equally spaced points on the unit circle has perimeter 1 3. A calculator shows that this number is approximately 6.1. The perimeter of this dodecagon should be just somewhat less than the circumference of the unit circle, as can be seen in the figure that is the solution to Exercise 8. The unit circle has circumference π, which is approximately 6.8. Thus the perimeter of the dodecagon (approximately 6.1) is somewhat close to the circumference of the circle (approximately 6.8) as we indeed expect from the figure.
45 Instructor s Solutions Manual, Section 6.1 Problem Explain why a regular polygon with n sides whose vertices are n equally spaced points on the unit circle has area n sin π n. solution Consider a regular polygon with n sides whose vertices are n equally spaced points on the unit circle. Draw a radius from the origin to each vertex, partitioning the polygon into n isosceles triangles. Each of these n isosceles triangles has two sides of length 1 (the radii), with angle π n between these two sides (because the entire circle, which has angle π, has been partitioned into n equal angles). Thus the area of each of the n isosceles triangles is sin π n, which equals 1 sin π n. Because the regular polygon with n sides is composed of n of these triangles, the area of the polygon equals n sin π n.
46 Instructor s Solutions Manual, Section 6.1 Problem Explain why the result stated in the previous problem implies that sin π n π n for large positive integers n. solution Suppose n is a large positive integer. Then a regular polygon with n sides whose vertices are n equally spaced points on the unit circle fills up almost all the area inside the unit circle. Thus the area of this polygon, which from the previous problem equals n sin π n, should be approximately the same as the area inside the unit circle, which equals π. In other words, we have n sin π n π. Multiplying both sides by n gives the approximation sin π n π n.
47 Instructor s Solutions Manual, Section 6.1 Problem Choose three large values of n, and use a calculator to verify that sin π n π n for each of those three large values of n. solution The table below gives the values of sin π n and π n significant digits for n = 1000, n = 10000, and n = : to six n sin π n π n As can be seen from the table above, we have sin π n π n for these three large values of n, with the approximation more accurate for larger values of n.
48 Instructor s Solutions Manual, Section 6.1 Problem Show that each edge of a regular polygon with n sides whose vertices are n equally spaced points on the unit circle has length cos π n. solution Consider a regular polygon with n sides whose vertices are n equally spaced points on the unit circle, with one of the vertices at the point (1, 0). The next vertex in the counterclockwise direction has coordinates ( cos π n, sin π ) n, because the radius ending at that vertex has angle π n with the positive horizontal axis. Thus the length of this edge of the polygon (and hence of each edge of the polygon) is the distance between the points (1, 0) and ( π cos n, sin π ) n. We can compute that distance using the usual formula for the distance between two points. Thus the length of each edge of this polygon is given by the following formula: (1 π ) cos n + sin π n 1 = cos π n + cos π n + sin π n = cos π n
49 Instructor s Solutions Manual, Section 6.1 Problem Explain why a regular polygon with n sides, each with length s, has area n sin π n 4(1 cos π n ) s. solution A regular polygon with n sides, with the vertices equally spaced on the unit circle, has area n sin π n (from Problem 45) and has sides of length cos π n (from Problem 48). Thus to get a regular polygon with sides having length s, we horizontally and vertically stretch this polygon with vertices on the unit circle by a factor of s. cos π n By the Area Stretch Theorem (Section 4.), this changes the area by a factor of ( ) s. cos π n Thus a regular polygon with n sides, each with length s, has area which equals ( ) n (sin π n ) s, cos π n
50 Instructor s Solutions Manual, Section 6.1 Problem 49 n sin π n 4(1 cos π n ) s.
51 Instructor s Solutions Manual, Section 6.1 Problem Verify that for n = 4, the formula given by the previous problem reduces to the usual formula for the area of a square. solution If n = 4, then n sin π n 4 sin π 4(1 cos π n ) s = 4(1 cos π ) s = s, which is the usual formula for the area of a square with sides of length s.
52 Instructor s Solutions Manual, Section 6.1 Problem Explain why a regular polygon with n sides whose vertices are n equally spaced points on the unit circle has perimeter n cos π n. solution By Problem 48, we know that each side of a regular polygon with n sides whose vertices lie on the unit circle has length cos π n. The perimeter of the polygon is n times the length of each side. Thus the perimeter of this polygon is n cos π n.
53 Instructor s Solutions Manual, Section 6.1 Problem 5 5. Explain why the result stated in the previous problem implies that n cos π n π for large positive integers n. solution If n is a large positive integer, then the perimeter of regular polygon with n sides whose vertices lie on the unit circle is approximately equal to the circumference of the unit circle, which equals π. In other words, if n is a large positive integer then n cos π n π.
54 Instructor s Solutions Manual, Section 6.1 Problem Choose three large values of n, and use a calculator to verify that n cos π n π for each of those three large values of n. solution The table below gives the values of n cos π n and π to six significant digits for n = 100, n = 1000, and n = 10000: n n cos π n π As can be seen from the table above, we have n cos π n π for these three large values of n, with the approximation more accurate for larger values of n.
55 Instructor s Solutions Manual, Section 6.1 Problem Show that cos π n 1 π n if n is a large positive integer. solution Suppose n is a large positive integer. Then we know from Problem 5 that n cos π n π. Square both sides of the approximation above and then divide by n to get cos π n 4π n. Now divide both sides by and then solve for cos π n, getting cos π n 1 π n.
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