solution guide for chapter 6: the geometry of right...

Solution Guide for Chapter 6:The Geometryof Right Triangles

6.1 THE THEOREM OF PYTHAGORAS

E-1. Another demonstration:

(a) Each triangle has area12ab, so the sum of the areas of the triangles is 4

(12ab

). One

square has area b2, and the other has area a2. So the total area is b2 + a2 + 4(

12ab

).

(b) Each triangle has area12ab, so the sum of the areas of the triangles is 4

(12ab

). The

square has area c2. So the total area is c2 + 4(

12ab

).

(c) Equating the two expressions gives

b2 + a2 + 4(

12ab

)= c2 + 4

(12ab

),

and canceling the common term of 4(

12ab

)leaves

a2 + b2 = c2.

E-2. Yet another demonstration:

(a) The distance between the parallel sides is a + b, and the average of the lengths of

those sides is12

(a + b). The area is the product of these two numbers, so it equals

12

(a + b) (a + b) =12

(a + b)2 .

(b) Two of the triangles have area12ab, and the third has area

12c2.

(c) Expanding the expression in Part (a) using the rules of algebra gives

12(a2 + 2ab + b2

).

490 Solution Guide for Chapter 6

By Part (b) the area of the trapezoid is

2(

12ab

)+

12c2.

Equating these two expressions gives

12(a2 + 2ab + b2

)= 2

(12ab

)+

12c2.

Canceling the common term of 2(

12ab

)and then multiplying through by 2 gives

a2 + b2 = c2.

E-3. Other demonstrations: Answers will vary.

S-1. A right triangle: Now 92+402 = 412, so the sides satisfy the equation in the Pythagorean

theorem. Hence this is a right triangle.

S-2. A right triangle: Now 362+772 = 852, so the sides satisfy the equation in the Pythagorean

theorem. Hence this is a right triangle.

S-3. Find the missing side: By the Pythagorean theorem the hypotenuse has length√32 + 72 =

√58 = 7.62.

S-4. Find the missing side: By the Pythagorean theorem the hypotenuse has length√42 + 92 =

√97 = 9.85.

S-5. Find the missing side: By the Pythagorean theorem the other leg has length√72 − 42 =

√33 = 5.74.

S-6. Find the missing side: By the Pythagorean theorem the other leg has length√92 − 52 =

√56 = 7.48.

S-7. Finding area: The triangle has area12(4× 5) = 10.

S-8. Finding area: The triangle has area12(7× 10) = 35.

SECTION 6.1 The Theorem of Pythagoras 491

S-9. Finding perimeter: By the Pythagorean theorem the hypotenuse has length√

22 + 32 =√

13. Thus the perimeter is 2 + 3 +√

13 = 8.61.

S-10. Finding perimeter: By the Pythagorean theorem the other leg has length√

32 − 22 =√

5. Thus the perimeter is 2 + 3 +√

5 = 7.24.

S-11. A funny right triangle? By the Pythagorean theorem the other leg would have length√

32 − 32 = 0. No such triangle is possible.

1. Area and perimeter: By the Pythagorean theorem a =√

82 − 52 =√

39 and b =√

72 − 52 =√

24. Thus the area is

12× 5×

(√39 +

√24)

= 27.86,

and the perimeter is

7 + 8 +√

39 +√

24 = 26.14.

2. Area and perimeter: By the Pythagorean theorem a =√

132 − 102 =√

69 and b =√

172 − 102 =√


12× 10×

(√69 +

√189)

= 110.27,


17 + 13 +√

69 +√

189 = 52.05.

3. Area and perimeter: By the Pythagorean theorem the indicated altitude has length√

132 − 62 =√

133. Applying the Pythagorean theorem again gives b =√

172 − 133 =√


12×√

133×(√

156 + 6)

= 106.62,


17 + 13 +√

156 + 6 = 48.49.


4. Area and perimeter: By the Pythagorean theorem the indicated altitude has length√

92 − 52 =√

56. Applying the Pythagorean theorem again gives b =√

172 − 56 =√

233. Thus the area is12×√

56×(√

233 + 5)

= 75.82,


17 + 9 +√

233 + 5 = 46.26.

5. A formula for Pythagorean triples:

(a) To verify the formula (a2−b2)2 +(2ab)2 = (a2 +b2)2, we expand the terms on each

side to see if they are equal. The left-hand side of the formula is

(a2 − b2)2 + (2ab)2 = (a2)2 − 2(a2)(b2) + (b2)2 + 4a2b2

= a4 − 2a2b2 + b4 + 4a2b2

= a4 + 2a2b2 + b4.

The right-hand side of the formula is

(a2 + b2)2 = a4 + 2a2b2 + b4.

Thus the two expressions are equal.

(b) Using the formula in Part (a), we see that if r = a2 − b2, s = 2ab, and t = a2 + b2

for some a and b, then (r, s, t) is a Pythagorean triple. For example, if b = 1 and

a is any positive integer larger than 1, then we see that (a2 − 1, 2a, a2 + 1) is a

Pythagorean triple, and so we can generate many such Pythagorean triples, as

shown below.

a Pythagorean triple2 (3,4,5)3 (8,6,10)4 (15,8,17)5 (24,10,26)6 (35,12,37)


6. Verifying the area formula: Refer to the accompanying figure.

Altitud

e

Base

Extension

of Base

We use b to denote the length of the base, h the length of the altitude, and t the length of

the extension of the base. Then the large triangle is a right triangle with base of length

t + b, so its area is12h(t + b). Inside the large triangle is another right triangle, one with

base of length t and area12ht. Subtracting these two areas gives the area of the original

triangle:12h(t + b)− 1

2ht =

12hb.

This is the desired formula.

7. An isosceles triangle: The altitude divides the isosceles triangle into two right trian-

gles, and the altitude is a leg of each of these. Also, each of these has hypotenuse a

and another leg on lengthb

2, so (by the Pythagorean theorem) the altitude has length√

a2 − b2

4. Then the area of the isosceles triangle is

12b

√a2 − b2

4,

which is the desired formula.

8. An equilateral triangle: Every equilateral triangle is isosceles, so the formula in Ex-

ercise 7 applies. In this case the base length b equals a, so the equilateral triangle has

areaa

2

√a2 − a2

4=

a

2

√3a2

4=√

3a2

4,

as desired.


9. Area and perimeter:

(a) By the Pythagorean theorem the hypotenuse has length√

a2 + b2. Hence the

perimeter is a + b +√

a2 + b2.

(b) Since the area is 4, we get12ab = 4.

Solving for b gives

b =8a.

Then the perimeter is

a +8a

+

√a2 +

(8a

)2

.

(c) We want to know when the perimeter is 15. So we need to solve the equation

a +8a

+

√a2 +

(8a

)2

= 15.

We do this using the crossing-graphs method. From the figure on the left below

we see that there are two crossing points and thus two possible values for a. (We

used a horizontal span from 0 to 10 and a vertical span from 0 to 25.)

From this figure we read a value of a = 1.16. It turns out that the second crossing

point occurs at a = 6.87. Both of these values of a produce a triangle with a

perimeter of 15 units.

(d) We want to find when the graph we have made reaches a minimum. This is shown

in the figure on the right above, where we see that the minimum perimeter of 9.66

occurs at a = 2.83.


10. Area and perimeter:

(a) By the Pythagorean theorem the hypotenuse has length√

a2 + b2. Hence the

perimeter is a + b +√

a2 + b2.

(b) Since the area is 10, we get12ab = 10.

Solving for b gives

b =20a

.

Then the perimeter is

a +20a

+

√a2 +

(20a

)2

.

(c) We want to know when the perimeter is 25. So we need to solve the equation

a +20a

+

√a2 +

(20a

)2

= 25.

We do this using the crossing-graphs method. From the figure on the left below

we see that there are two crossing points and thus two possible values for a. (We

used a horizontal span from 0 to 15 and a vertical span from 0 to 30.)

From this figure we read a value of a = 1.73. It turns out that the second crossing

point occurs at a = 11.57. Both of these values of a produce a triangle with a


(d) We want to find when the graph we have made reaches a minimum. This is shown

in the figure on the right above, where we see that the minimum perimeter of 15.27

occurs at a = 4.47.

11. A funny triangle? By the solution to Part 2(d) of the example, the minimum possible

perimeter for such a triangle is 10.80. Hence it is not possible to make such a triangle

with a perimeter of 5 units.


12. A funny triangle? According to Part 2(b) of the example, the perimeter of such a trian-

gle is

a +10a

+

√a2 +

(10a

)2

.

Here a denotes a leg. So we want to know if there is a solution to the equation

a +10a

+

√a2 +

(10a

)2

= 500.

We do this using the crossing-graphs method. In the figure below we see one crossing

point. (We used a horizontal span from 0 to 1 and a vertical span from 0 to 600.)

The corresponding value is a = 0.04. It turns out that there is another crossing point,

one that gives the value a = 249.98. Both of these values of a produce a triangle with a


13. Finding a side: Let a denote the length of the other leg. By the Pythagorean theorem

the hypotenuse has length√

64 + a2. Thus in terms of a the perimeter is

8 + a +√

64 + a2.

So we want to solve the equation

8 + a +√

64 + a2 = 20.

We do this using the crossing-graphs method. In the figure below we used a horizontal

span from 0 to 20 and a vertical span from 0 to 30.


We see that the corresponding value is a = 3.33. This is the length of the other leg.

14. Finding the lengths of legs:

(a) By the Pythagorean theorem b =√

82 − a2 =√

64− a2.

(b) The perimeter is 8 + a + b, and by Part (a) this equals

8 + a +√

64− a2.

(c) By Part (b) we want to solve the equation

8 + a +√

64− a2 = 17.

We do this using the crossing-graphs method. From the figure below we see that

there are two crossing points and thus two possible values for a. (We used a hori-

zontal span from 0 to 8 and a vertical span from 0 to 25.)

From this figure we read a value of a = 1.07. The corresponding value of b is

then b =√

64− 1.072 = 7.93. (It turns out that the second crossing point occurs at

a = 7.93—no surprise—and of course the corresponding value of b is then 1.07.)



152 − a2 =√

225− a2.

(b) The perimeter is 15 + a + b, and by Part (a) this equals

15 + a +√

225− a2.

(c) By Part (b) we want to solve the equation

15 + a +√

225− a2 = 35.

We do this using the crossing-graphs method. From the figure below we see that

there are two crossing points and thus two possible values for a. (We used a hori-

zontal span from 0 to 15 and a vertical span from 0 to 40.)


From this figure we read a value of a = 6.46. The corresponding value of b is then

b =√

225− 6.462 = 13.54. (It turns out that the second crossing point occurs at

a = 13.54—no surprise—and of course the corresponding value of b is then 6.46.)


(a) In general, the sum of the lengths of the legs must be larger than the hypotenuse

because the shortest distance between two points is along a line. Here is another

way to see this: If the hypotenuse has length c and the legs have lengths a and b,

then

(a + b)2 = a2 + b2 + 2ab > a2 + b2 = c2,

where the last equality holds by the Pythagorean theorem. Thus a + b > c. Of

course, in this exercise c = 15. Hence the sum of the lengths of the legs must be at

least 15.

(b) No, because by Part (a) the sum of the lengths of the legs must be larger than 15,

so the perimeter must be larger than 30.

(c) In general, the hypotenuse is the longest side by the Pythagorean theorem: If the

hypotenuse has length c and the legs have lengths a and b, then c2 = a2 + b2 > a2,

so c is larger than a. Of course, in this exercise c = 15. Hence the length of each

leg is at most 15.

(d) No, because by Part (c) the length of each leg is less than 15, so the perimeter is

less than 45.

17. Maximizing area:


202 − a2 =√

400− a2.

(b) The area is12ab, and by Part (a) this equals

12a√

400− a2.


(c) By Part (b) we want to find where the function

12a√

400− a2

attains its maximum value. From the figure below we see that this occurs where

a = 14.14. (We used a horizontal span from 0 to 20 and a vertical span from 0 to

110.) The corresponding value of b is then b =√

400− 14.142 = 14.14. It makes

sense that the maximum area occurs when the two legs are of equal length.

18. Maximizing area:


652 − a2 =√

4225− a2.

(b) The area is12ab, and by Part (a) this equals

12a√

4225− a2.

(c) By Part (b) we want to find where the function

12a√

4225− a2

attains its maximum value. From the figure below we see that this occurs where

a = 45.96. (We used a horizontal span from 0 to 65 and a vertical span from 0 to

1200.) The corresponding value of b is then b =√

4225− 45.962 = 45.96. It makes

sense that the maximum area occurs when the two legs are of equal length.


19. Cartography: The line segment from our observation point to the peak is the hy-

potenuse of a right triangle. The hypotenuse has length 2.2 miles, and one leg (from

our observation point to the base) has length 1.6 miles. By the Pythagorean theorem

the other leg is√

2.22 − 1.62 = 1.51 miles in length. Thus the height of the wall is about

1.5 miles.

20. Cartography: The line segment from our observation point to the peak is the hy-

potenuse of a right triangle. The hypotenuse has length 2 miles, and one leg (the wall

itself) has length 0.16 mile. By the Pythagorean theorem the other leg is√

22 − 0.162 =

1.99 miles in length. That is the distance from our observation point to the base of the

wall.

6.2 ANGLES

E-1. The Pythagorean theorem:

(a) The leg of length c1 in the smaller triangle corresponds to the leg of length a in the

larger triangle. The hypotenuse of the smaller triangle on the left has length a, and

the hypotenuse of the larger triangle has length c1 + c2. By similarity we have

c1

a=

a

c1 + c2.

(b) The leg of length c2 in the smaller triangle corresponds to the leg of length b in the

larger triangle. The hypotenuse of the smaller triangle on the right has length b,

and the hypotenuse of the larger triangle has length c1 + c2. By similarity we have

c2

b=

b

c1 + c2.

(c) In the equation from Part (a) we cross-multiply and recall that c1 + c2 = c. The

result is

a2 = cc1.

Doing the same thing in the equation from Part (b) yields

b2 = cc2.

Adding these equations gives

a2 + b2 = cc1 + cc2 = c(c1 + c2) = c2,

as desired.

SECTION 6.2 Angles 501

E-2. Isosceles right triangle:

(a) Consider the smaller triangle on the left. It shares an angle with the larger triangle

(the angle whose vertex is at the lower left). Both triangles have a right angle. This

means that the third pair of angles must also be equal, so the triangles are similar.

The same argument applies to the smaller triangle on the right.

(b) The hypotenuse of the larger triangle has length 2, and the hypotenuse of the

smaller triangle has length x. For the angle in common the corresponding lengths

are x for the larger triangle and 1 for the smaller triangle. By similarity we have

x

1=

2x

,

and so x2 = 2. Hence x =√

2.

S-1. Conversion: This is 3× 180π

= 171.89 degrees.

S-2. Conversion: This isπ

7× 180

π= 25.71 degrees.

S-3. Conversion: This is 72× π

180= 1.26 radians.

S-4. Conversion: This is π × π

180= 0.05 radian.

S-5. Angle sum: The radian measure of the third angle is

π − π

9− 0.8 = 1.99.

S-6. Angle sum: The degree measure of the third angle is

180− 20− 40 = 120.

S-7. Angle sum: Becauseπ

4is 45 degrees, the degree measure of the third angle is

180− 45− 30 = 105.

The radian measure is 105× π

180= 1.83 .

S-8. Angle sum: Because 20 degrees is 20× π

180=

π

9radian, the radian measure of the third

angle is π− π

9− 0.3 = 2.49. The degree measure is 142.81. (Here we converted to degree

measure before rounding the radian measure.)


S-9. Angle sum: Because 70 degrees is 70 × π

180=

7π

18radian, the radian measure of the

third angle is

π − 7π

18− 0.4 = 1.52.

The degree measure is 87.08. (Here we converted to degree measure before rounding

the radian measure.)

S-10. Angle sum: All three angles are equal, and they add up to 180 degrees, so the degree

measure of an angle is1803

= 60.

S-11. Angle sum: All three angles are equal, and they add up to π radians, so the radian

measure of an angle isπ

3= 1.05.

1. Area and arc length: The arc length isπ

6× 5 = 2.62 units. The area is

π

6× 52

2= 6.54

square units.

2. Area and arc length: The arc length is 1.4× 9 = 12.6 units. The area is 1.4× 92

2= 56.7

square units.

3. Area and arc length: The arc length is25× π × 10

180= 4.36 units. The area is

25× π × 102

360=

21.82 square units.


180= 0.89 unit. The area is

17× π × 32

360=

1.34 square units.

5. Area and arc length: Let a denote the radian measure of the angle and r the radius. We

know that ar = 5 and thatar2

2= 9. Dividing the second equation by the first gives

r

2=

95

. Thus the radius is r = 2 × 95

= 3.6 units. Putting this result into the equation

ar = 5 gives that 3.6a = 5, so the angle is a =5

3.6= 1.39 radians.

6. Area and arc length: Let a denote the radian measure of the angle and r the radius. We

know that ar = 12 and thatar2

2= 17. Dividing the second equation by the first gives

r

2=

1712

. Thus the radius is r = 2 × 1712

=176

= 2.83 units. Putting this result into

the equation ar = 12 gives that the angle is a = 12 ÷ 176

=7217

radians. This is 242.66

degrees.

7. Doubling: If we consider the circle formed by the bottom of the pie, then the bigger

piece represents twice the area of the smaller. Now in general the area is proportional

to the angle, so doubling the area requires doubling the angle. Hence the bigger piece

(that of your friend) has twice the central angle of the smaller one (yours). To be explicit,

let a denote the radian measure of the smaller angle and A that of the bigger angle. Let

SECTION 6.2 Angles 503

r denote the radius. The area for the smaller angle isar2

2, and the area for the bigger

angle isAr2

2. The second of these is twice the first, so

Ar2

2= 2 × ar2

2. Cancelling

common terms gives A = 2a.

8. More on doubling: If we consider the circle formed by the bottom of each pie, then the

circle for the larger pie has twice the diameter of the smaller. Let r denote the radius

of the smaller circle. Then the radius of the larger circle is 2r. Let a denote the radian

measure corresponding to a piece. (This is the same for both circles because the number

of pieces is the same.) The area for a piece of the smaller pie isar2

2, and the area for a

piece of the larger pie is

a(2r)2

2=

a× 22r2

2= 4× ar2

2.

Thus the pieces of the larger pie are 4 times those of the smaller.

9. Latitude: The distance from Fort Worth to Wichita is the arc length cut by an angle of

37 − 32 = 5 degrees out of a circle of radius 3950 miles. That length is5× π × 3950

180=

344.70 miles.

10. More latitude: Now we are given the arc length (1180 miles), and we need to find the

angle. The radius is still 3950 miles. If the angle is d degrees thend× π × 3950

180= 1180.

Solving for d gives d = 1180÷ π × 3950180

= 17.12 degrees. Thus the latitude of Winnipeg

is 32 + 17.12 = 49.12, or about 49, degrees north.

11. Similarity: By similarity we haveB

8=

318

and thus B = 8 × 318

= 1.33. In a similar

way we findc

2=

183

and thus c = 2× 183

= 12.

12. Similarity: By similarity we haveB

10=

320

and thus B = 10 × 320

= 1.5. In a similar

way we findc

2=

203

and thus c = 2× 203

= 13.33.

13. Similarity: By similarity we havey

1=

1x

and thus y =1x

.

14. Similarity: By similarity we havey

1=

x

yand thus y2 = x. Hence y =

√x.

15. Grads: Because 90 degrees represents one-quarter of a circle, the grad measure of this

angle is 400 × 14

= 100. Because π radians represents one-half of a circle, the grad

measure of this angle is 400 × 12

= 200. An angle of 32 degrees represents32360

of a

circle, so the grad measure of this angle is 400× 32360

= 35.56.


16. Mils:

(a) Now 1 mil is1

1000radian, and 1 radian is

180π

degrees. Hence 1 mil is

11000

× 180π

= 0.057 degree.

(b) Let r denote the distance, in yards, from the soldier to the target. We know that an

angle of 1 mil cuts an arc length of 1 yard out of a circle of radius r. Because 1 mil

is1

1000radian, the formula for arc length says

11000

× r = 1.

Hence the distance is r = 1000 yards.

(c) Now there are 1000 mils in a radian, and there are 2π radians in a circle. Hence

there are 1000× 2π = 6283.19 mils in a circle.

17. Shadows: There are two right triangles here. One has the pole and its shadow as legs,

and the other has the tree and its shadow. Because the position of the sun is the same

for both shadows, these right triangles are similar. Let t denote the height of the tree.

Then by similarityt

10=

216

.

Thus

t = 10× 216

= 35,

so the tree is 35 feet tall.

18. Ladder: There are two right triangles to consider here. One has the ladder as hy-

potenuse and a 5-foot horizontal segment as a leg. The other has the lower 12 feet

of the ladder as hypotenuse and for a leg the horizontal segment from the base of the

ladder to the point beneath the rung. We denote by d the length of this latter leg. Be-

cause these right triangles have the angle of the ladder with the horizontal in common,

the triangles are similar. By similarity

d

5=

1220

.

Thus

d = 5× 1220

= 3.

Hence the distance from the rung to the wall is 5− 3 = 2 feet.

SECTION 6.3 Right Angle Trigonometry 505

19. An angle and a circle: By similarity we have

|AB||AD|

=|AE||AC|

.

Cross-multiplying gives |AB| × |AC| = |AD| × |AE|, as desired.

20. Area equals arc length: Let r denote the radius of the circle. We know that for every

angle of radian measure a the corresponding area and arc length are the same, so

ar =ar2

2.

This implies that

r =r2

2,

and solving for r gives r = 2.

6.3 RIGHT TRIANGLE TRIGONOMETRY

E-1. Other trigonometric functions of special angles: We know that

sin 30◦ =12

and

cos 30◦ =√

32

.

Hence

tan 30◦ =sin 30◦

cos 30◦=

12÷

(√3

2

)=

1√3,

cot 30◦ =1

tan 30◦= 1÷

(1√3

)=√

3,

sec 30◦ =1

cos 30◦= 1÷

(√3

2

)=

2√3,

and

csc 30◦ =1

sin 30◦= 1÷

(12

)= 2.

E-2. Other trigonometric functions of special angles: We know that

sinπ

4= cos

π

4=

1√2.

Hence

tanπ

4=

sin π4

cos π4

=1√2÷(

1√2

)= 1,


cotπ

4=

1tan π

4

= 1÷ 1 = 1,

secπ

4=

1cos π

4

= 1÷(

1√2

)=√

2,

and

cscπ

4=

1sin π

4

= 1÷(

1√2

)=√

2.

E-3. The 15-degree angle:

(a) Because the angle marked s has degree measure 15, the angle of the big triangle

that includes s has degree measure 60 + 15 = 75. Now we show that the angle

marked t has measure 75 degrees. In addition to the angle marked t, the big tri-

angle has a 30-degree angle, and also the angle of measure 75 degrees that we just

considered. Because the angle sum of the triangle is 180 degrees, the angle marked

t has measure 180−30−75 = 75 degrees. Here is another way to see the same thing:

The adjoined smaller right triangle has acute angles s and t. Because the angle sum

of the triangle is 180 degrees, the angle marked t has measure 180 − 90 − 15 = 75

degrees.

(b) The sides opposite the 75-degree angles in the big triangle are the same, so A+√

3 =

2, and thus A = 2−√

3.

(c) We apply the Pythagorean theorem to the adjoined smaller right triangle:

B2 = 12 + A2,

so by Part (b) we have

B2 = 1 + (2−√

3)2 = 1 + 4− 4√

3 + 3 = 8− 4√

3.

Thus

B =√

8− 4√

3.

(d) In the adjoined smaller right triangle, the angle marked s has degree measure 15,

the side opposite s is A, and the hypotenuse is B. Hence

sin 15◦ =Opposite

Hypotenuse=

A

B=

2−√

3√8− 4

√3,

as desired.


E-4. An angle ofπ

8radian:

(a) Because the angle marked s has radian measureπ

8, the angle of the big triangle that

includes s has radian measureπ

4+

π

8=

3π

8. Now we show that the angle marked

t has radian measure3π

8. In addition to the angle marked t, the big triangle has an

angle of radian measureπ

4, and also the angle of radian measure

3π

8that we just

considered. Because the angle sum of the triangle is π radians, the angle marked t

has radian measure π − π

4− 3π

8=

3π

8. Here is another way to see the same thing:

The adjoined smaller right triangle has acute angles s and t. Because the angle sum

of the triangle is π radians, the angle marked t has radian measure π− π

2− π

8=

3π

8.

(b) The sides opposite the angles in the big triangle of radian measure3π

8are the same,

so A + 1 =√

2, and thus A =√

2− 1.

(c) We apply the Pythagorean theorem to the adjoined smaller right triangle:

B2 = 12 + A2,

so by Part (b) we have

B2 = 1 + (√

2− 1)2 = 1 + 2− 2√

2 + 1 = 4− 2√

2.

Thus

B =√

4− 2√

2.

(d) In the adjoined smaller right triangle, the angle marked s has radian measureπ

8,

the side opposite s is A, and the hypotenuse is B. Hence

sinπ

8=

OppositeHypotenuse

=A

B=

√2− 1√

4− 2√

2,

as desired.

S-1. Calculating trigonometric functions: Now

sin θ =Opposite

Hypotenuse=

1517

= 0.88,

cos θ =Adjacent

Hypotenuse=

817

= 0.47,

and

tan θ =OppositeAdjacent

=158

= 1.875, or about 1.88.



sin θ =Opposite

Hypotenuse=

3034

= 0.88,

cos θ =Adjacent

Hypotenuse=

1634

= 0.47,

and


=3016

= 1.875, or about 1.88.


sin θ =Opposite

Hypotenuse=

2425

= 0.96,

cos θ =Adjacent

Hypotenuse=

725

= 0.28,

and


=247

= 3.43.


sin θ =Opposite

Hypotenuse=

7074

= 0.95,

cos θ =Adjacent

Hypotenuse=

2474

= 0.32,

and


=7024

= 2.92.

S-5. Calculating trigonometric functions: By the Pythagorean theorem the length of the

hypotenuse is√

72 + 82 =√

113. Now

sin θ =Opposite

Hypotenuse=

8√113

= 0.75,

cos θ =Adjacent

Hypotenuse=

7√113

= 0.66,

and


=87

= 1.14.



hypotenuse is√

52 + 122 =√

169 = 13. Now

sin θ =Opposite

Hypotenuse=

1213

= 0.92,

cos θ =Adjacent

Hypotenuse=

513

= 0.38,

and


=125

= 2.40.


opposite side is√

122 − 92 =√

63. Now

sin θ =Opposite

Hypotenuse=√

6312

= 0.66,

cos θ =Adjacent

Hypotenuse=

912

= 0.75,

and


=√

639

= 0.88.


opposite side is√

502 − 152 =√

2275. Now

sin θ =Opposite

Hypotenuse=√

227550

= 0.95,

cos θ =Adjacent

Hypotenuse=

1550

= 0.30,

and


=√

227515

= 3.18.

S-9. Getting a length: If θ is the angle and H is the hypotenuse then

0.11 = sin θ =Opposite

Hypotenuse=

8H

,

so

H =8

0.11= 72.73.


S-10. Getting a length: If θ is the angle and H is the hypotenuse then

0.71 = cos θ =Adjacent

Hypotenuse=

8H

,

so

H =8

0.71= 11.27.

S-11. Getting an angle: We want to find θ so that sin θ = 0.53. We use the crossing-graphs

method in degree mode with a horizontal span from 0 to 90 degrees and a vertical span

from 0 to 1. We see from the graph below that θ = 32.01 degrees.

S-12. Getting an angle: We want to find θ so that tan θ = 4. We use the crossing-graphs

method in radian mode with a horizontal span from 0 toπ

2radians and a vertical span

from 0 to 5. We see from the graph below that θ = 1.33 radians.

S-13. Other trigonometric functions: Now

sec 30◦ =1

cos 30◦= 1.15,

csc 30◦ =1

sin 30◦= 2,

and

cot 30◦ =1

tan 30◦= 1.73.

Here we used the calculator to find the cosine, sine, and tangent. The exact values can

be found in the tables before the Enrichment Exercises.


S-14. Other trigonometric functions: Now

secπ

4=

1cos π

4

= 1.41,

cscπ

4=

1sin π

4

= 1.41,

and

cotπ

4=

1tan π

4

= 1.

Here we used the calculator to find the cosine, sine, and tangent. The exact values can

be found in the tables before the Enrichment Exercises.

1. Calculating height: If the man sits 331 horizontal feet from the base of a wall and he

must incline his eyes at an angle of 16.2 degrees to look at the top of the wall, then the

ground and the wall form two legs of a right triangle with angle θ = 16.2 degrees, and

the adjacent side has length 331 feet. Since we want to find the length of the opposite

side, we use the tangent (as it is the ratio of opposite to adjacent):


tan 16.2◦ =Opposite

331Opposite = 331 tan 16.2◦ = 96.16.

Thus the wall is 96.16 feet tall.

2. Calculating height: If the man sits 222 horizontal feet from the base of a wall and he

must incline his eyes at an angle of 6.4 degrees to look at the top of the wall, then the

ground and the wall form two legs of a right triangle with angle θ = 6.4 degrees, and

the adjacent side has length 222 feet. Since we want to find the length of the opposite

side, we use the tangent (as it is the ratio of opposite to adjacent):


tan 6.4◦ =Opposite

222Opposite = 222 tan 6.4◦ = 24.90.

Thus the wall is 24.90 feet tall.

3. Calculating an angle: If the man sits 270 horizontal feet from the base of a wall which

is 76 feet high, then the ground and the wall form two legs of a right triangle with

adjacent side of length 270 feet and opposite side of length 76 feet. Since we want to


find the angle θ at which he must incline his eyes to view the top of the wall, we use

the tangent (as it is the ratio of opposite to adjacent):


=76270

.

We solve this using the crossing-graphs method. We use a horizontal span from −90 to

90 degrees and a vertical span from 0 to 1. We see from the graph below that θ = 15.72

degrees. In radians the solution is 0.27.

4. Calculating an angle: If the wall is 38 feet high, the man sits on the ground, and the

distance from the man to the top of the wall is 83 feet, then the ground and the wall

form two legs of a right triangle with opposite side of length 38 feet and a hypotenuse

of length 83 feet. Since we want to find the angle θ at which he must incline his eyes to

view the top of the wall, we use the sine (as it is the ratio of opposite to hypotenuse):

sin θ =Opposite

Hypotenuse

=3883

.

We solve this using the crossing-graphs method. We use a horizontal span from 0 to 90

degrees and a vertical span from 0 to 1. We see from the graph below that θ = 27.25



5. Calculating distance: The horizontal distance of the man from the wall in Exercise 4 is

the length of the adjacent side of the triangle. Since we already know the lengths of the

other two sides, as well as the angle θ, there are many ways to calculate the horizontal

distance. One way is to use cosine (as it is the ratio of adjacent to hypotenuse):

cos θ =Adjacent

Hypotenuse

cos 27.25◦ =Adjacent

83Adjacent = 83 cos 27.25◦ = 73.79.

Thus the horizontal distance of the man from the wall is 73.79 feet.

6. Calculating distance: If the man sits 130 horizontal feet from the base of a wall and he

must incline his eyes at an angle of 13 degrees to look at the top of the wall, then the

ground and the wall form two legs of a right triangle with angle θ = 13 degrees, and

the adjacent side has length 130 feet. Since we want to find the distance from the man

directly to the top of the wall, that is, the length of the hypotenuse, we use the cosine

(as it is the ratio of adjacent to hypotenuse):

cos θ =Adjacent

Hypotenuse

cos 13◦ =130

Hypotenuse.

Hence

Hypotenuse =130

cos 13◦= 133.42,

so the distance from the man directly to the top of the wall is 133.42 feet.

7. Calculating distance: If the man sits 18 horizontal feet from the base of a wall and he

must incline his eyes at an angle of 21 degrees to look at the top of the wall, then the

ground and the wall form two legs of a right triangle with angle θ = 21 degrees, and

the adjacent side has length 18 feet. Since we want to find the distance from the man

directly to the top of the wall, that is, the length of the hypotenuse, we use the cosine

(as it is the ratio of adjacent to hypotenuse):

cos θ =Adjacent

Hypotenuse

cos 21◦ =18

Hypotenuse.

Hence

Hypotenuse =18

cos 21◦= 19.28,

so the distance from the man directly to the top of the wall is 19.28 feet.


8. The 3-4-5 right triangle:

(a) In a 3-4-5 right triangle, the legs have lengths 3 and 4, while the hypotenuse has

length 5. One acute angle has an opposite side of length 3 and hypotenuse of

length 5, so its sine is35

= 0.6, while the other acute angle has an opposite side of

length 4 and hypotenuse of length 5, so its sine is45

= 0.8.

(b) To find the two acute angles of the 3-4-5 right triangle, by Part (a) we need only

solve the equations sin θ = 0.6 and sin θ = 0.8. The corresponding solutions (found

using crossing graphs, for example) are θ = 36.87 degrees and θ = 53.13 degrees,

respectively. In radians the angles are 0.64 and 0.93, respectively.

9. Finding the angle: We want to find θ so that cot θ = 5, or

cos θ

sin θ= 5.

We solve this using the crossing-graphs method. We use a horizontal span from 0 to 90

degrees and a vertical span from 0 to 10. We see from the graph below that θ = 11.31

degrees.

10. A building: If you must incline your transit at an angle of 20 degrees to look at the

top of the building, which is 150 feet taller than the transit, then the ground and the

building form two legs of a right triangle with angle θ = 20 degrees, and the opposite

side has length 150 feet. Since we want to find the horizontal distance from your transit

to the building, that is, the length of the adjacent side, we use the tangent (as it is the

ratio of opposite to adjacent):


tan 20◦ =150

Adjacent.

Hence

Adjacent =150

tan 20◦= 412.12,

so the horizontal distance from your transit to the building is 412.12 feet.


11. The width of a river: If you must rotate your transit through an angle of 12 degrees

to point toward the second tree, and the distance between the trees is 35 yards, then

the two trees and the transit form the vertices of a right triangle, with the right angle

at the first tree. This is a right angle because the transit-to-first-tree line is north-south,

while the line between the trees is east-west. The transit-to-first-tree and the first-tree-

to-second-tree sides form two legs of a right triangle with angle θ = 12 degrees, and the

opposite side has length 35 yards. Since we want to find the distance from your transit

to the first tree, that is, the length of the adjacent side, we use the tangent (as it is the

ratio of opposite to adjacent):


tan 12◦ =35

Adjacent.

Hence

Adjacent =35

tan 12◦= 164.66,

so the distance from your transit to the first tree, that is, the width of the river, is 164.66

yards, or about 165 yards.

12. A cannon: The distance downrange a cannonball strikes the ground is given by

m2 sin (2t)g

,

where, in our case, g = 32 and m = 300, and we want the cannonball to strike the

ground 1000 feet downrange. Thus we want to find t such that3002 sin(2t)

32= 1000.

This equation can be solved using the crossing-graphs method. We use a horizontal

span from 0 to 90 degrees and a vertical span from 0 to 2000. We see from the graph

below that one solution is an angle of t = 10.41 degrees. Another angle that could be

used is t = 79.59 degrees. In radians these are 0.18 and 1.39, respectively.


13. Dallas to Fort Smith: Since Dallas is 190 miles due south of Oklahoma City, and Fort

Smith is 140 miles due east of Oklahoma City, the triangle Dallas to Oklahoma City to

Fort Smith to Dallas is a right triangle with the right angle at Oklahoma City, and the

airplane’s flight path forms the hypotenuse.

(a) The tangent of the angle that the flight path makes with Interstate 35 is given byOppositeAdjacent

=140190

= 0.74.

(b) The angle θ that the flight path makes with Interstate 35 has the property that

tan θ =140190

. Solving, for example using crossing graphs, shows that θ = 36.38


(c) The distance that the airplane flies is the length of the hypotenuse. Using cosine,

for example, we can calculate that length:

cos θ =Adjacent

Hypotenuse

cos 36.38◦ =190

Hypotenuse.

Hence

Hypotenuse =190

cos 36.38◦= 236.00,

so the length of the flight path is 236 miles.

14. Intensity of sunlight:

(a) If the angle θ is 35 degrees, then the intensity is reduced by a factor of sin θ =

sin 35◦ = 0.57.

(b) If the intensity is reduced by a factor of 0.3, then the angle θ of the incident rays

satisfies the relation sin θ = 0.3. Solving (for example using crossing graphs), we

find that θ = 17.46 degrees.

15. Grasping prey: As the diagram shows, the triangle ABC is a right triangle with the

right angle at B. The diameter |BC| is the opposite side from the angle θ, while the

length |AC| is the hypotenuse. Sine relates the lengths of the opposite side and the

hypotenuse:

sin θ =Opposite

Hypotenuse

sin θ =|BC||AC|

sin θ × |AC| = |BC|.

Thus the formula is |BC| = |AC| sin θ.


16. Getting the tangent from the sine:

(a) We are given that

23

= sin θ =Opposite

Hypotenuse,

so one such triangle has a side of length 2 opposite θ and a hypotenuse of length

3, as in the figure below. All such triangles are similar to this one.

3 2

0

(b) By the Pythagorean theorem, the length of the adjacent side is√

32 − 22 =√

5 =

2.24.

(c) Now


=2√5

= 0.89.

17. Getting the cosine from the tangent:

(a) We are given that

51

= tan θ =OppositeAdjacent

,

so one such triangle has a side of length 5 opposite θ and a side of length 1 adjacent

to θ, as in the figure below. All such triangles are similar to this one.


5

0

1

(b) By the Pythagorean theorem, the length of the hypotenuse is√

52 + 12 =√

26 =

5.10.

(c) Now

cos θ =Adjacent

Hypotenuse=

1√26

= 0.20.

18. Getting the sine from the cotangent:

(a) We are given that23

= cot θ =AdjacentOpposite

,



3

0

2


32 + 22 =√

13 =

3.61.


(c) Now

sin θ =Opposite

Hypotenuse=

3√13

= 0.83.

19. Dispersal method:

(a) As indicated in the figure, v is the length of the side opposite the angle θ and d is

the length of the hypotenuse. Thus sin θ =v

dand so v = d sin θ.

(b) If the distance the organisms move is 30 centimeters, then d = 30, and so the

formula from Part (a) becomes v = 30 sin θ. Thus if the angle θ is 15 degrees,

then v = 30 sin 15◦ = 7.76 centimeters, while if the angle is 30 degrees, then v =

30 sin 30◦ = 15 centimeters.

(c) If the distance the organisms move is d = 30 centimeters and the change in eleva-

tion is v = 20 centimeters, then, by the formula from Part (a), sin θ =2030

. Solving

for θ, for example using crossing graphs, we obtain θ = 41.81 degrees.

20. Jumping locust: We are given that g = 9.8, so the formula for the horizontal distance

that an animal jumps is

d =m2 sin 2θ

9.8.

(a) If a locust jumps at an angle of θ = 55 degrees for a horizontal distance d of 0.8

meter, then 0.8 =m2 sin(2× 55◦)

9.8, so m2 sin(2 × 55◦) = 0.8 × 9.8. Hence m2 =

0.8× 9.8sin(2× 55◦)

. Finally, the initial velocity is

m =

√0.8× 9.8

sin(2× 55◦)= 2.89 meters per second.

(b) If the distance that an animal jumps is 1 meter and the initial velocity is 3.2 meters

per second, then 1 =3.22 sin(2θ)

9.8, and so 9.8 = 3.22 sin(2θ), or

sin(2θ) =9.83.22

.

Solving for θ, for example by using crossing graphs, we find that θ = 36.57 de-

grees.


Chapter 6 Review Exercises

1. Area and perimeter: By the Pythagorean theorem

a =√

72 − 32 =√

40 = 6.32

and

b =√

102 − a2 =√

100− 40 =√

60 = 7.75.

Thus the area is12×√

40×(3 +

√60)

= 33.98,


10 + 7 + 3 +√

60 = 27.75.

2. Maximizing area and perimeter:

(a) By the Pythagorean theorem the length of the other leg is√

302 − a2 =√

900− a2.

(b) The perimeter is a + 30 +√

900− a2.

(c) The area is12a√

900− a2.

(d) We want to find when the graph of the perimeter function from Part (b) reaches a

maximum. We made the graph using a horizontal span from 0 to 30 and a vertical

span from 0 to 100. In the figure below we see that the maximum perimeter of

72.43 occurs at a = 21.21.

Chapter 6 Review Exercises 521

(e) We want to find when the graph of the area function from Part (c) reaches a maxi-

mum. We made the graph using a horizontal span from 0 to 30 and a vertical span

from 0 to 300. In the figure below we see that the maximum area of 225 occurs at

a = 21.21.

3. Lengths of legs:

(a) In general, the sum of the lengths of the legs must be larger than the hypotenuse

because the shortest distance between two points is along a line. Here is another

way to see this: If the hypotenuse has length c and the legs have lengths a and b,

then

(a + b)2 = a2 + b2 + 2ab > a2 + b2 = c2,

where the last equality holds by the Pythagorean theorem. Thus a + b > c. Of

course, in this exercise c = 20. Hence the sum of the lengths of the legs must be at

least 20.

(b) In general, the hypotenuse is the longest side by the Pythagorean theorem: If the

hypotenuse has length c and the legs have lengths a and b, then c2 = a2 + b2 > a2,

so c is larger than a. Of course, in this exercise c = 20. Hence the sum of the

lengths of the legs is less than twice the length of the hypotenuse, so is less than

40.

(c) Yes. By the Pythagorean theorem the length of the other leg is√

202 − 12 =√

399 = 19.97.

(d) No: By the reasoning in Part (b), the hypotenuse is longer than each leg.

4. Pythagorean triples:

(a) Such a triangle is similar to the 3-4-5 right triangle, so it has a right angle.

(b) There are many possibilities, including (3,4,5), (5,12,13), and (8,15,17). These can

be found by trial and error, or by using Exercise 5 in Section 6.1.


(c) We would need to find a right triangle whose legs are whole numbers a and b and

whose hypotenuse is 3. Then a and b would have to be less than 3, so each would

have to be 1 or 2. No such choice satisfies the equation a2 + b2 = 32 from the

Pythagorean theorem.


180= 2.79 inches. The area is

20× π × 82

360=

11.17 square inches.

6. Angle sum of a triangle:

(a) The angle sum of a triangle is 180 degrees or π radians.

(b) The sum of the two acute angles is 180− 90 = 90 degrees orπ

2radians.

(c) Now 1 radian is180π

= 57.30 degrees, and that is the degree measure of the angle

at vertex B. The degree measure of the angle at vertex C is 180− 60− 180π

= 62.70.

(d) Now 60 degrees is 60× π

180=

π

3= 1.05 radians, and that is the radian measure of

the angle at vertex A. The radian measure of the angle at vertex C is π − π

3− 1 =

1.09. (This can also be found by converting the answer from Part (c) to radians.)

7. Similar triangles: We use vertical bars to denote the length of a segment.

(a) These are right triangles, and they have the angle at E in common, so they are

similar.

(b) By similarity we have|AE||CE|

=|AB||CD|

.

Thus|AE|13

=1712

,

so |AE| = 13× 1712

= 18.42.

(c) Now

|AC| = |AE| − |CE| = 18.42− 13 = 5.42.

We find |BD| by first finding |BE|. By similarity we have

|BE||DE|

=|AB||CD|

.

Thus|BE|

5=

1712

,

so |BE| = 5× 1712

= 7.08. Hence

|BD| = |BE| − |DE| = 7.08− 5 = 2.08.


The length of BE can also be found using by the Pythagorean theorem. There is

small variation due to rounding.

(d) The quadrilateral is a right trapezoid, and its area is

12(|AB|+ |CD|)|BD| = 1

2(17 + 12)× 2.08 = 30.16.

Here is another way to do this: The area of the quadrilateral is the difference be-

tween the areas of triangle ABE and triangle CDE. There is small variation due to

rounding.

8. Ladder against a wall:

(a) We consider the right triangle with the ladder as hypotenuse, a 17-foot vertical

segment as a leg, and a 6-foot horizontal segment as a leg. By the Pythagorean

theorem the length of the hypotenuse is√

172 + 62 = 18.03. Thus the ladder is

about 18 feet long.

(b) In addition to the right triangle from Part (a), there is another right triangle to

consider here. It has the lower 4 feet of the ladder as hypotenuse and for a leg the

vertical segment from the rung to the ground. We denote by d the length in feet of

this latter leg. Because these right triangles have the angle of the ladder with the

horizontal in common, the triangles are similar. By similarity

d

17=

418

.

Thus

d = 17× 418

= 3.78.

Hence the rung is about 3.78 feet above the ground. The answer will vary slightly

if we use the length of the ladder from Part (a) without rounding.

9. Trigonometric functions:

(a) By the Pythagorean theorem the length of the hypotenuse is√102 + 72 =

√149 = 12.21.

(b) Now

sin θ =Opposite

Hypotenuse=

7√149

= 0.57,

cos θ =Adjacent

Hypotenuse=

10√149

= 0.82,

and


=710

= 0.70.


(c) Now

csc θ =1

sin θ=√

1497

= 1.74,

sec θ =1

cos θ=√

14910

= 1.22,

and

cot θ =1

tan θ=

107

= 1.43.

Using the rounded values from Part (b) leads to some variation.

10. Calculating vertical distances:

(a) The ground and the cliff form the two legs of a right triangle with angle θ = 13

degrees, and the adjacent side has length 1620 feet. The tangent function is the

ratio of opposite to adjacent, so it involves the height of the cliff and the distance

from the base of the cliff.

(b) Let c denote the height of the cliff in feet. By the reasoning in Part (a),

tan 13◦ =c

1620,

so

c = 1620 tan 13◦ = 374.01.

Thus the height of the cliff is 374.01 feet.

(c) We want the length of the hypotenuse of the right triangle described in Part (a).

Let H denote that length in feet. We use the cosine function because it involves

the adjacent side and the hypotenuse. We have

cos 13◦ =Adjacent

Hypotenuse=

1620H

,

so

H =1620

cos 13◦= 1662.61.

Thus the distance from the person to the top of the cliff is 1662.61 feet. This can

also be found using the Pythagorean theorem and the answer from Part (b).

11. Calculating horizontal distances:

(a) The height above the road of the mountain’s peak is 14,610− 5220 = 9390 feet.

(b) The road and the mountain form the two legs of a right triangle with angle θ = 9.5

degrees, and by Part (a) the opposite side has length 9390 feet. We use the tangent

function because it is the ratio of opposite to adjacent, so it involves the height of


the peak above the road and the distance from the base of the mountain. Let d

denote that distance in feet. Then

tan 9.5◦ =9390

d,

so

d =9390

tan 9.5◦= 56,112.43.

Thus the distance from the base of the mountain is about 56,112 feet, or 10.63 miles.

12. Finding an angle: We want to find θ so that tan θ = 2.3. We solve this using the

crossing-graphs method. We use a horizontal span from 0 to 90 using degrees and a

vertical span from 0 to 5. We see from the graph below that θ = 66.50 degrees.

13. Getting the sine and cosine from the tangent:

(a) We are given that13

= tan θ =OppositeAdjacent

,



1

0

3


12 + 32 =√

10 =

3.16.

(c) Now

sin θ =Opposite

Hypotenuse=

1√10

= 0.32,


and

cos θ =Adjacent

Hypotenuse=

3√10

= 0.95.

14. Jumping:

(a) If an animal jumps at an angle of θ = 50 degrees and the horizontal distance d is

1.1 meters, then 1.1 =1

9.8m2 sin(2× 50◦), so m2 =

1.1× 9.8sin(2× 50◦)

. Hence the initial

velocity is

m =

√1.1× 9.8

sin(2× 50◦)= 3.31 meters per second.

(b) If the distance that an animal jumps is 2 meters and the initial velocity is 6 meters

per second, then 2 =1

9.862 sin(2θ), and so

sin(2θ) =2× 9.8

62.

This equation can be solved using the crossing-graphs method. We use a horizon-

tal span from 0 to 90 degrees and a vertical span from 0 to 1. We see from the graph

below that there are two solutions and that one solution is an angle of θ = 16.49

degrees. The other solution is an angle of θ = 73.51 degrees.

(c) If the distance that an animal jumps is 2 meters and the initial velocity is 4 meters

per second, then 2 =1

9.842 sin(2θ), and so

sin(2θ) =2× 9.8

42.

Now the right-hand side of this equation is2× 9.8

42= 1.225, which is greater than

1. Because the sine function is never greater than 1, there is no solution to this

equation. (This can also be seen from a graph.) Hence there are no such angles.


15. The 5-12-13 right triangle:

(a) Because 52 + 122 = 132, the 5-12-13 triangle is a right triangle.

(b) In a 5-12-13 right triangle, the legs have lengths 5 and 12, while the hypotenuse

has length 13. One acute angle has an opposite side of length 5 and hypotenuse of

length 13, so its sine is513

= 0.38, and the other acute angle has an opposite side

of length 12 and hypotenuse of length 13, so its sine is1213

= 0.92.

(c) To find the two acute angles of the 5-12-13 right triangle, by Part (b) we need only

solve the equations sin θ =513

and sin θ =1213

. The corresponding solutions (found

using crossing graphs, for example) are θ = 22.62 degrees and θ = 67.38 degrees,

respectively.

(d) In radians the angles from Part (c) are 0.39 and 1.18.

solution guide for chapter 6: the geometry of right...

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