Download Solution Guide for Chapter 6: The Geometry of Right Triangles

Solution Guide for Chapter 6:
The Geometry
of Right Triangles
6.1 THE THEOREM OF PYTHAGORAS
E-1. Another demonstration:
1
ab . One
2 1
2
2
2
2
square has area b , and the other has area a . So the total area is b + a + 4
ab .
2
1
1
(b) Each triangle has area ab, so the sum of the areas of the triangles is 4
ab . The
2
2
1
2
2
square has area c . So the total area is c + 4
ab .
2
1
(a) Each triangle has area ab, so the sum of the areas of the triangles is 4
2
(c) Equating the two expressions gives
1
1
ab = c2 + 4
ab ,
2
2
1
and canceling the common term of 4
ab leaves
2
b2 + a2 + 4
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
1
those sides is (a + b). The area is the product of these two numbers, so it equals
2
1
1
2
(a + b) (a + b) = (a + b) .
2
2
1
1
(b) Two of the triangles have area ab, and the third has area c2 .
2
2
(c) Expanding the expression in Part (a) using the rules of algebra gives
1 2
a + 2ab + b2 .
2
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Solution Guide for Chapter 6
By Part (b) the area of the trapezoid is
1
1
2
ab + c2 .
2
2
Equating these two expressions gives
1 2
1
1
2
a + 2ab + b = 2
ab + c2 .
2
2
2
1
Canceling the common term of 2
ab and then multiplying through by 2 gives
2
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
p
√
32 + 72 = 58 = 7.62.
S-4. Find the missing side: By the Pythagorean theorem the hypotenuse has length
p
√
42 + 92 = 97 = 9.85.
S-5. Find the missing side: By the Pythagorean theorem the other leg has length
p
√
72 − 42 = 33 = 5.74.
S-6. Find the missing side: By the Pythagorean theorem the other leg has length
p
√
92 − 52 = 56 = 7.48.
1
S-7. Finding area: The triangle has area (4 × 5) = 10.
2
1
S-8. Finding area: The triangle has area (7 × 10) = 35.
2
SECTION 6.1
491
The Theorem of Pythagoras
√
S-9. Finding perimeter: By the Pythagorean theorem the hypotenuse has length
√
√
13. Thus the perimeter is 2 + 3 + 13 = 8.61.
S-10. Finding perimeter: By the Pythagorean theorem the other leg has length
√
√
5. Thus the perimeter is 2 + 3 + 5 = 7.24.
√
22 + 32 =
32 − 22 =
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 =
√
√
72 − 52 = 24. Thus the area is
√
82 − 52 =
√
39 and b =
√
√ 1
×5×
39 + 24 = 27.86,
2
and the perimeter is
7+8+
√
39 +
√
24 = 26.14.
2. Area and perimeter: By the Pythagorean theorem a =
√
√
172 − 102 = 189. Thus the area is
√
132 − 102 =
√
69 and b =
√
√
1
× 10 ×
69 + 189 = 110.27,
2
and the perimeter is
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 =
√
156. Thus the area is
√
1 √
156 + 6 = 106.62,
× 133 ×
2
and the perimeter is
17 + 13 +
√
156 + 6 = 48.49.
492
Solution Guide for Chapter 6
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 is
√
1 √
× 56 ×
233 + 5 = 75.82,
2
and the perimeter is
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 + 4a2 b2
= a4 − 2a2 b2 + b4 + 4a2 b2
= a4 + 2a2 b2 + b4 .
The right-hand side of the formula is
(a2 + b2 )2 = a4 + 2a2 b2 + 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
2
3
4
5
6
Pythagorean triple
(3,4,5)
(8,6,10)
(15,8,17)
(24,10,26)
(35,12,37)
SECTION 6.1
The Theorem of Pythagoras
493
Altitude
6. Verifying the area formula: Refer to the accompanying figure.
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
1
t + b, so its area is h(t + b). Inside the large triangle is another right triangle, one with
2
1
base of length t and area ht. Subtracting these two areas gives the area of the original
2
triangle:
1
1
1
h(t + b) − ht = hb.
2
2
2
This is the desired formula.
7. An isosceles triangle: The altitude divides the isosceles triangle into two right triangles, and the altitude is a leg of each of these. Also, each of these has hypotenuse a
b
and another leg on length , so (by the Pythagorean theorem) the altitude has length
2
r
b2
a2 − . Then the area of the isosceles triangle is
4
r
1
b2
b a2 − ,
2
4
which is the desired formula.
8. An equilateral triangle: Every equilateral triangle is isosceles, so the formula in Exercise 7 applies. In this case the base length b equals a, so the equilateral triangle has
area
a
2
as desired.
r
a2
a
a2 −
=
4
2
r
3a2
=
4
√
3a2
,
4
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Solution Guide for Chapter 6
9. Area and perimeter:
(a) By the Pythagorean theorem the hypotenuse has length
√
perimeter is a + b + a2 + b2 .
p
a2 + b2 . Hence the
(b) Since the area is 4, we get
1
ab = 4.
2
Solving for b gives
b=
8
.
a
Then the perimeter is
8
a+ +
a
s
a2 +
2
8
.
a
(c) We want to know when the perimeter is 15. So we need to solve the equation
s
2
8
8
a + + a2 +
= 15.
a
a
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.
SECTION 6.1
The Theorem of Pythagoras
495
10. Area and perimeter:
(a) By the Pythagorean theorem the hypotenuse has length
√
perimeter is a + b + a2 + b2 .
p
a2 + b2 . Hence the
(b) Since the area is 10, we get
1
ab = 10.
2
Solving for b gives
b=
20
.
a
Then the perimeter is
s
20
a+
+
a
a2
+
20
a
2
.
(c) We want to know when the perimeter is 25. So we need to solve the equation
20
a+
+
a
s
a2
+
20
a
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
perimeter of 25 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 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.
496
Solution Guide for Chapter 6
12. A funny triangle? According to Part 2(b) of the example, the perimeter of such a triangle is
10
a+
+
a
s
a2 +
10
a
2
.
Here a denotes a leg. So we want to know if there is a solution to the equation
s
2
10
10
a+
+ a2 +
= 500.
a
a
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
perimeter of 500 units.
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
p
64 + a2 .
8+a+
So we want to solve the equation
8+a+
p
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.
SECTION 6.1
The Theorem of Pythagoras
497
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+
p
64 − a2 .
(c) By Part (b) we want to solve the equation
8+a+
p
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 horizontal 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.)
15. Finding the lengths of legs:
(a) By the Pythagorean theorem b =
√
152 − a2 =
√
225 − a2 .
(b) The perimeter is 15 + a + b, and by Part (a) this equals
15 + a +
p
225 − a2 .
(c) By Part (b) we want to solve the equation
15 + a +
p
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 horizontal span from 0 to 15 and a vertical span from 0 to 40.)
498
Solution Guide for Chapter 6
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.)
16. Finding the 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 = 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:
(a) By the Pythagorean theorem b =
√
202 − a2 =
√
1
(b) The area is ab, and by Part (a) this equals
2
1 p
a 400 − a2 .
2
400 − a2 .
SECTION 6.1
The Theorem of Pythagoras
499
(c) By Part (b) we want to find where the function
1 p
a 400 − a2
2
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:
(a) By the Pythagorean theorem b =
√
652 − a2 =
√
4225 − a2 .
1
(b) The area is ab, and by Part (a) this equals
2
1 p
a 4225 − a2 .
2
(c) By Part (b) we want to find where the function
1 p
a 4225 − a2
2
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.
500
Solution Guide for Chapter 6
19. Cartography: The line segment from our observation point to the peak is the hypotenuse 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 hypotenuse 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
2
x
= ,
1
x
and so x2 = 2. Hence x =
S-1. Conversion: This is 3 ×
√
2.
180
= 171.89 degrees.
π
180
π
×
= 25.71 degrees.
7
π
π
S-3. Conversion: This is 72 ×
= 1.26 radians.
180
π
= 0.05 radian.
S-4. Conversion: This is π ×
180
S-2. Conversion: This is
S-5. Angle sum: The radian measure of the third angle is
π−
π
− 0.8 = 1.99.
9
S-6. Angle sum: The degree measure of the third angle is
180 − 20 − 40 = 120.
S-7. Angle sum: Because
π
is 45 degrees, the degree measure of the third angle is
4
180 − 45 − 30 = 105.
π
= 1.83 .
180
π
π
S-8. Angle sum: Because 20 degrees is 20 ×
= radian, the radian measure of the third
180
9
π
angle is π − − 0.3 = 2.49. The degree measure is 142.81. (Here we converted to degree
9
measure before rounding the radian measure.)
The radian measure is 105 ×
502
Solution Guide for Chapter 6
S-9. Angle sum: Because 70 degrees is 70 ×
π
7π
=
radian, the radian measure of the
180
18
third angle is
π−
7π
− 0.4 = 1.52.
18
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
180
measure of an angle is
= 60.
3
S-11. Angle sum: All three angles are equal, and they add up to π radians, so the radian
π
measure of an angle is = 1.05.
3
1. Area and arc length: The arc length is
π
52
π
× 5 = 2.62 units. The area is ×
= 6.54
6
6
2
square units.
2. Area and arc length: The arc length is 1.4 × 9 = 12.6 units. The area is 1.4 ×
92
= 56.7
2
square units.
3. Area and arc length: The arc length is
25 × π × 102
25 × π × 10
= 4.36 units. The area is
=
180
360
21.82 square units.
4. Area and arc length: The arc length is
17 × π × 32
17 × π × 3
= 0.89 unit. The area is
=
180
360
1.34 square units.
5. Area and arc length: Let a denote the radian measure of the angle and r the radius. We
ar2
= 9. Dividing the second equation by the first gives
know that ar = 5 and that
2
r
9
9
= . Thus the radius is r = 2 × = 3.6 units. Putting this result into the equation
2
5
5
5
ar = 5 gives that 3.6a = 5, so the angle is a =
= 1.39 radians.
3.6
6. Area and arc length: Let a denote the radian measure of the angle and r the radius. We
ar2
know that ar = 12 and that
= 17. Dividing the second equation by the first gives
2
r
17
17
17
=
. Thus the radius is r = 2 ×
=
= 2.83 units. Putting this result into
2
12
12
6
17
72
the equation ar = 12 gives that the angle is a = 12 ÷
=
radians. This is 242.66
6
17
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
ar2
, and the area for the bigger
2
2
Ar
Ar2
ar2
angle is
. The second of these is twice the first, so
= 2×
. Cancelling
2
2
2
common terms gives A = 2a.
r denote the radius. The area for the smaller angle is
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
ar2
of pieces is the same.) The area for a piece of the smaller pie is
, and the area for a
2
piece of the larger pie is
a(2r)2
a × 22 r2
ar2
=
=4×
.
2
2
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
5 × π × 3950
37 − 32 = 5 degrees out of a circle of radius 3950 miles. That length is
=
180
344.70 miles.
10. More latitude: Now we are given the arc length (1180 miles), and we need to find the
d × π × 3950
angle. The radius is still 3950 miles. If the angle is d degrees then
= 1180.
180
π × 3950
Solving for d gives d = 1180 ÷
= 17.12 degrees. Thus the latitude of Winnipeg
180
is 32 + 17.12 = 49.12, or about 49, degrees north.
B
3
3
11. Similarity: By similarity we have
=
and thus B = 8 ×
= 1.33. In a similar
8
18
18
c
18
18
way we find =
and thus c = 2 ×
= 12.
2
3
3
B
3
3
12. Similarity: By similarity we have
=
and thus B = 10 ×
= 1.5. In a similar
10
20
20
c
20
20
way we find =
and thus c = 2 ×
= 13.33.
2
3
3
y
1
1
= and thus y = .
1
x
x
√
y
x
14. Similarity: By similarity we have = and thus y 2 = x. Hence y = x.
1
y
13. Similarity: By similarity we have
15. Grads: Because 90 degrees represents one-quarter of a circle, the grad measure of this
1
angle is 400 ×
= 100. Because π radians represents one-half of a circle, the grad
4
1
32
measure of this angle is 400 × = 200. An angle of 32 degrees represents
of a
2
360
32
circle, so the grad measure of this angle is 400 ×
= 35.56.
360
504
Solution Guide for Chapter 6
16. Mils:
(a) Now 1 mil is
1
180
radian, and 1 radian is
degrees. Hence 1 mil is
1000
π
1
180
×
= 0.057 degree.
1000
π
(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
1
is
radian, the formula for arc length says
1000
1
× r = 1.
1000
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 similarity
t
21
=
.
10
6
Thus
t = 10 ×
21
= 35,
6
so the tree is 35 feet tall.
18. Ladder: There are two right triangles to consider here. One has the ladder as hypotenuse 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. Because these right triangles have the angle of the ladder with the horizontal in common,
the triangles are similar. By similarity
12
d
=
.
5
20
Thus
d=5×
12
= 3.
20
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|
|AE|
=
.
|AD|
|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
r=
r2
,
2
This implies that
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◦ =
and
√
◦
cos 30 =
Hence
1
2
3
.
2
√ !
1
3
=√ ,
2
3
√
1
1
= 3,
cot 30◦ =
=1÷ √
◦
tan 30
3
√ !
3
2
1
=√ ,
sec 30◦ =
=1÷
◦
2
cos 30
3
1
sin 30◦
= ÷
tan 30◦ =
cos 30◦
2
and
csc 30◦ =
1
=1÷
sin 30◦
1
= 2.
2
E-2. Other trigonometric functions of special angles: We know that
sin
Hence
π
π
1
= cos = √ .
4
4
2
sin π4
π
1
tan =
π = √ ÷
4
cos 4
2
1
√
2
= 1,
506
Solution Guide for Chapter 6
π
1
=
= 1 ÷ 1 = 1,
4
tan π4
√
π
1
1
sec =
=1÷ √
= 2,
4
cos π4
2
cot
and
csc
π
1
=
=1÷
4
sin π4
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 triangle 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:
B 2 = 12 + A2 ,
so by Part (b) we have
B 2 = 1 + (2 −
Thus
√
√
√
3)2 = 1 + 4 − 4 3 + 3 = 8 − 4 3.
q
√
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
√
Opposite
A
2− 3
sin 15 =
=
=p
√ ,
Hypotenuse
B
8−4 3
◦
as desired.
SECTION 6.3
E-4. An angle of
Right Angle Trigonometry
507
π
radian:
8
π
(a) Because the angle marked s has radian measure , the angle of the big triangle that
8
π
π
3π
includes s has radian measure + =
. Now we show that the angle marked
4
8
8
3π
t has radian measure
. In addition to the angle marked t, the big triangle has an
8
π
3π
angle of radian measure , and also the angle of radian measure
that we just
4
8
considered. Because the angle sum of the triangle is π radians, the angle marked t
π
3π
3π
has radian measure π − −
=
. Here is another way to see the same thing:
4
8
8
The adjoined smaller right triangle has acute angles s and t. Because the angle sum
π π
3π
of the triangle is π radians, the angle marked t has radian measure π − − =
.
2 8
8
3π
(b) The sides opposite the angles in the big triangle of radian measure
are the same,
8
√
√
so A + 1 = 2, and thus A = 2 − 1.
(c) We apply the Pythagorean theorem to the adjoined smaller right triangle:
B 2 = 12 + A2 ,
so by Part (b) we have
√
√
√
B 2 = 1 + ( 2 − 1)2 = 1 + 2 − 2 2 + 1 = 4 − 2 2.
Thus
q
√
B = 4 − 2 2.
(d) In the adjoined smaller right triangle, the angle marked s has radian measure
the side opposite s is A, and the hypotenuse is B. Hence
√
2−1
π
Opposite
A
sin =
=
=p
√ ,
8
Hypotenuse
B
4−2 2
as desired.
S-1. Calculating trigonometric functions: Now
sin θ =
Opposite
15
=
= 0.88,
Hypotenuse
17
cos θ =
Adjacent
8
=
= 0.47,
Hypotenuse
17
and
tan θ =
Opposite
15
=
= 1.875, or about 1.88.
Adjacent
8
π
,
8
508
Solution Guide for Chapter 6
S-2. Calculating trigonometric functions: Now
sin θ =
Opposite
30
=
= 0.88,
Hypotenuse
34
cos θ =
16
Adjacent
=
= 0.47,
Hypotenuse
34
and
tan θ =
Opposite
30
=
= 1.875, or about 1.88.
Adjacent
16
S-3. Calculating trigonometric functions: Now
sin θ =
24
Opposite
=
= 0.96,
Hypotenuse
25
cos θ =
Adjacent
7
=
= 0.28,
Hypotenuse
25
and
tan θ =
24
Opposite
=
= 3.43.
Adjacent
7
S-4. Calculating trigonometric functions: Now
sin θ =
70
Opposite
=
= 0.95,
Hypotenuse
74
cos θ =
Adjacent
24
=
= 0.32,
Hypotenuse
74
and
tan θ =
Opposite
70
=
= 2.92.
Adjacent
24
S-5. Calculating trigonometric functions: By the Pythagorean theorem the length of the
√
√
hypotenuse is 72 + 82 = 113. Now
sin θ =
Opposite
8
= 0.75,
=√
Hypotenuse
113
cos θ =
Adjacent
7
= 0.66,
=√
Hypotenuse
113
and
tan θ =
Opposite
8
= = 1.14.
Adjacent
7
SECTION 6.3
Right Angle Trigonometry
509
S-6. Calculating trigonometric functions: By the Pythagorean theorem the length of the
√
√
hypotenuse is 52 + 122 = 169 = 13. Now
sin θ =
Opposite
12
=
= 0.92,
Hypotenuse
13
cos θ =
5
Adjacent
=
= 0.38,
Hypotenuse
13
and
tan θ =
Opposite
12
=
= 2.40.
Adjacent
5
S-7. Calculating trigonometric functions: By the Pythagorean theorem the length of the
√
√
opposite side is 122 − 92 = 63. Now
√
Opposite
63
= 0.66,
sin θ =
=
12
Hypotenuse
cos θ =
Adjacent
9
=
= 0.75,
Hypotenuse
12
and
tan θ =
Opposite
=
Adjacent
√
63
= 0.88.
9
S-8. Calculating trigonometric functions: By the Pythagorean theorem the length of the
√
√
opposite side is 502 − 152 = 2275. Now
Opposite
sin θ =
=
Hypotenuse
cos θ =
√
2275
= 0.95,
50
Adjacent
15
=
= 0.30,
Hypotenuse
50
and
Opposite
tan θ =
=
Adjacent
√
2275
= 3.18.
15
S-9. Getting a length: If θ is the angle and H is the hypotenuse then
0.11 = sin θ =
8
Opposite
= ,
Hypotenuse
H
so
H=
8
= 72.73.
0.11
510
Solution Guide for Chapter 6
S-10. Getting a length: If θ is the angle and H is the hypotenuse then
0.71 = cos θ =
8
Adjacent
= ,
Hypotenuse
H
so
H=
8
= 11.27.
0.71
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
π
method in radian mode with a horizontal span from 0 to radians and a vertical span
2
from 0 to 5. We see from the graph below that θ = 1.33 radians.
S-13. Other trigonometric functions: Now
1
= 1.15,
cos 30◦
1
csc 30◦ =
= 2,
sin 30◦
sec 30◦ =
and
cot 30◦ =
1
= 1.73.
tan 30◦
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.
SECTION 6.3
Right Angle Trigonometry
511
S-14. Other trigonometric functions: Now
sec
π
1
=
= 1.41,
4
cos π4
csc
π
1
=
= 1.41,
4
sin π4
and
cot
1
π
=
= 1.
4
tan π4
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):
Opposite
Adjacent
Opposite
tan 16.2◦ =
331
Opposite = 331 tan 16.2◦ = 96.16.
tan θ
=
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):
Opposite
Adjacent
Opposite
tan 6.4◦ =
222
Opposite = 222 tan 6.4◦ = 24.90.
tan θ
=
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
512
Solution Guide for Chapter 6
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):
tan θ
=
=
Opposite
Adjacent
76
.
270
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
38
.
83
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
degrees. In radians the solution is 0.48.
SECTION 6.3
Right Angle Trigonometry
513
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):
Adjacent
Hypotenuse
Adjacent
cos 27.25◦ =
83
Adjacent = 83 cos 27.25◦ = 73.79.
cos θ
=
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
=
cos 13◦
=
Adjacent
Hypotenuse
130
.
Hypotenuse
Hence
Hypotenuse =
130
= 133.42,
cos 13◦
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
=
cos 21◦
=
Adjacent
Hypotenuse
18
.
Hypotenuse
Hence
Hypotenuse =
18
= 19.28,
cos 21◦
so the distance from the man directly to the top of the wall is 19.28 feet.
514
Solution Guide for Chapter 6
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
3
length 5, so its sine is = 0.6, while the other acute angle has an opposite side of
5
4
length 4 and hypotenuse of length 5, so its sine is = 0.8.
5
(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
= 5.
sin θ
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 θ
=
tan 20◦
=
Opposite
Adjacent
150
.
Adjacent
Hence
Adjacent =
150
= 412.12,
tan 20◦
so the horizontal distance from your transit to the building is 412.12 feet.
SECTION 6.3
Right Angle Trigonometry
515
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-treeto-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 θ
=
tan 12◦
=
Opposite
Adjacent
35
.
Adjacent
Hence
Adjacent =
35
= 164.66,
tan 12◦
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
3002 sin(2t)
ground 1000 feet downrange. Thus we want to find t such that
= 1000.
32
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.
516
Solution Guide for Chapter 6
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 by
Opposite
140
=
= 0.74.
Adjacent
190
(b) The angle θ that the flight path makes with Interstate 35 has the property that
140
tan θ =
. Solving, for example using crossing graphs, shows that θ = 36.38
190
degrees. In radians the solution is 0.64.
(c) The distance that the airplane flies is the length of the hypotenuse. Using cosine,
for example, we can calculate that length:
cos θ
=
cos 36.38◦
=
Adjacent
Hypotenuse
190
.
Hypotenuse
Hence
Hypotenuse =
190
= 236.00,
cos 36.38◦
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:
Opposite
Hypotenuse
|BC|
sin θ =
|AC|
sin θ × |AC| = |BC|.
sin θ
Thus the formula is |BC| = |AC| sin θ.
=
SECTION 6.3
Right Angle Trigonometry
517
16. Getting the tangent from the sine:
(a) We are given that
2
Opposite
= sin θ =
,
3
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
tan θ =
Opposite
2
= √ = 0.89.
Adjacent
5
17. Getting the cosine from the tangent:
(a) We are given that
5
Opposite
= tan θ =
,
1
Adjacent
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.
518
Solution Guide for Chapter 6
5
0
1
(b) By the Pythagorean theorem, the length of the hypotenuse is
√
52 + 12 =
√
26 =
5.10.
(c) Now
cos θ =
Adjacent
1
= √ = 0.20.
Hypotenuse
26
18. Getting the sine from the cotangent:
(a) We are given that
2
Adjacent
= cot θ =
,
3
Opposite
so one such triangle has a side of length 3 opposite θ and a side of length 2 adjacent
to θ, as in the figure below. All such triangles are similar to this one.
3
0
2
(b) By the Pythagorean theorem, the length of the hypotenuse is
3.61.
√
32 + 22 =
√
13 =
SECTION 6.3
Right Angle Trigonometry
519
(c) Now
sin θ =
Opposite
3
= √ = 0.83.
Hypotenuse
13
19. Dispersal method:
(a) As indicated in the figure, v is the length of the side opposite the angle θ and d is
v
the length of the hypotenuse. Thus sin θ = and so v = d sin θ.
d
(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 eleva20
tion is v = 20 centimeters, then, by the formula from Part (a), sin θ =
. Solving
30
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
m2 sin(2 × 55◦ )
, so m2 sin(2 × 55◦ ) = 0.8 × 9.8. Hence m2 =
meter, then 0.8 =
9.8
0.8 × 9.8
. Finally, the initial velocity is
sin(2 × 55◦ )
s
0.8 × 9.8
m=
= 2.89 meters per second.
sin(2 × 55◦ )
(b) If the distance that an animal jumps is 1 meter and the initial velocity is 3.2 meters
3.22 sin(2θ)
per second, then 1 =
, and so 9.8 = 3.22 sin(2θ), or
9.8
sin(2θ) =
9.8
.
3.22
Solving for θ, for example by using crossing graphs, we find that θ = 36.57 degrees.
520
Solution Guide for Chapter 6
Chapter 6 Review Exercises
1. Area and perimeter: By the Pythagorean theorem
a=
and
b=
Thus the area is
p
72 − 32 =
√
40 = 6.32
p
√
√
102 − a2 = 100 − 40 = 60 = 7.75.
√ 1 √
× 40 × 3 + 60 = 33.98,
2
and the perimeter is
10 + 7 + 3 +
√
60 = 27.75.
2. Maximizing area and perimeter:
(a) By the Pythagorean theorem the length of the other leg is
√
(b) The perimeter is a + 30 + 900 − a2 .
1 p
(c) The area is a 900 − a2 .
2
√
302 − a2 =
√
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.
521
Chapter 6 Review Exercises
(e) We want to find when the graph of the area function from Part (c) reaches a maximum. 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
√
399 = 19.97.
√
202 − 12 =
(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.
522
Solution Guide for Chapter 6
(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.
5. Area and arc length: The arc length is
20 × π × 82
20 × π × 8
= 2.79 inches. The area is
=
180
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 or
π
radians.
2
180
= 57.30 degrees, and that is the degree measure of the angle
π
180
at vertex B. The degree measure of the angle at vertex C is 180 − 60 −
= 62.70.
π
π
π
(d) Now 60 degrees is 60 ×
= = 1.05 radians, and that is the radian measure of
180
3
π
the angle at vertex A. The radian measure of the angle at vertex C is π − − 1 =
3
1.09. (This can also be found by converting the answer from Part (c) to radians.)
(c) Now 1 radian is
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|
|AB|
=
.
|CE|
|CD|
Thus
|AE|
17
=
,
13
12
so |AE| = 13 ×
17
= 18.42.
12
(c) Now
|AC| = |AE| − |CE| = 18.42 − 13 = 5.42.
We find |BD| by first finding |BE|. By similarity we have
|BE|
|AB|
=
.
|DE|
|CD|
Thus
|BE|
17
=
,
5
12
so |BE| = 5 ×
17
= 7.08. Hence
12
|BD| = |BE| − |DE| = 7.08 − 5 = 2.08.
Chapter 6 Review Exercises
523
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
1
1
(|AB| + |CD|)|BD| = (17 + 12) × 2.08 = 30.16.
2
2
Here is another way to do this: The area of the quadrilateral is the difference between 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
4
=
.
17
18
Thus
d = 17 ×
4
= 3.78.
18
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
p
√
102 + 72 = 149 = 12.21.
(b) Now
Opposite
7
= 0.57,
=√
Hypotenuse
149
Adjacent
10
cos θ =
= 0.82,
=√
Hypotenuse
149
sin θ =
and
tan θ =
Opposite
7
=
= 0.70.
Adjacent
10
524
Solution Guide for Chapter 6
(c) Now
√
149
= 1.74,
7
√
1
149
sec θ =
=
= 1.22,
cos θ
10
1
=
csc θ =
sin θ
and
cot θ =
1
10
=
= 1.43.
tan θ
7
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
1620
=
,
Hypotenuse
H
so
H=
1620
= 1662.61.
cos 13◦
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
525
Chapter 6 Review Exercises
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
= 56,112.43.
tan 9.5◦
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 that
Opposite
1
= tan θ =
,
3
Adjacent
so one such triangle has a side of length 1 opposite θ and a side of length 3 adjacent
to θ, as in the figure below. All such triangles are similar to this one.
1
0
3
(b) By the Pythagorean theorem, the length of the hypotenuse is
3.16.
(c) Now
sin θ =
Opposite
1
= √ = 0.32,
Hypotenuse
10
√
12 + 32 =
√
10 =
526
Solution Guide for Chapter 6
and
cos θ =
Adjacent
3
= √ = 0.95.
Hypotenuse
10
14. Jumping:
(a) If an animal jumps at an angle of θ = 50 degrees and the horizontal distance d is
1 2
1.1 × 9.8
1.1 meters, then 1.1 =
m sin(2 × 50◦ ), so m2 =
. Hence the initial
9.8
sin(2 × 50◦ )
velocity is
s
m=
1.1 × 9.8
= 3.31 meters per second.
sin(2 × 50◦ )
(b) If the distance that an animal jumps is 2 meters and the initial velocity is 6 meters
1 2
per second, then 2 =
6 sin(2θ), and so
9.8
sin(2θ) =
2 × 9.8
.
62
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 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
1 2
per second, then 2 =
4 sin(2θ), and so
9.8
sin(2θ) =
2 × 9.8
.
42
2 × 9.8
= 1.225, which is greater than
42
1. Because the sine function is never greater than 1, there is no solution to this
Now the right-hand side of this equation is
equation. (This can also be seen from a graph.) Hence there are no such angles.
Chapter 6 Review Exercises
527
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
5
length 13, so its sine is
= 0.38, and the other acute angle has an opposite side
13
12
of length 12 and hypotenuse of length 13, so its sine is
= 0.92.
13
(c) To find the two acute angles of the 5-12-13 right triangle, by Part (b) we need only
12
5
and sin θ =
. The corresponding solutions (found
solve the equations sin θ =
13
13
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.