MATH 103 HW5 Solutions
1. (a) Let a be the adjacent leg, b be the opposite leg, and c be the
b
2
hypotenuse. So tan θ = = . We may choose b = 2 and a = 3.
a
3
Now find c:
√
c2 = 32 + 22 = 13 =⇒ c = 13
√
13
2
sin θ = √
csc θ =
2
13
3
cos θ = √
sec θ =
13
√
13
3
2
3
cot θ =
3
2
(b) Let a be the adjacent leg, b be the opposite leg, and c be the
a
1
hypotenuse. So cot θ = = . We may choose a = 1 and b = 2.
b
2
Now find c:
√
c2 = 12 + 22 = 5 =⇒ c = 5
√
2
5
sin θ = √
csc θ =
2
5
tan θ =
1
cos θ = √
5
sec θ =
√
5
2
1
= 2 cot θ =
1
2
(c) Let a be the adjacent leg, b be the opposite leg, and c be the
c
5
hypotenuse. So csc θ = = 5 = . We may choose c = 5 and
b
1
b = 1. Now find a:
√
52 = a2 + 12 =⇒ a2 = 24 =⇒ a = 24
1
sin θ =
csc θ = 5
5
√
24
5
cos θ =
sec θ = √
5
24
tan θ =
1
tan θ = √
24
1
cot θ =
√
24
(d) Let a be the adjacent leg, b be the opposite leg, and c be the
5
c
= √ . We may choose c = 5 and
hypotenuse. So sec θ =
a
3
√
a = 3. Now find b:
√
√
52 = ( 3)2 + b2 =⇒ b2 = 22 =⇒ b = 22
√
22
5
sin θ =
csc θ = √
5
22
√
3
5
cos θ =
sec θ = √
5
3
√
√
22
3
cot θ = √
tan θ = √
3
22
2. Let a be the adjacent leg, b be the opposite leg, and c be the hypotenuse.
Then c = 10. Before we find a and b lets think about what values would
be reasonable for a and b. Since c is by definition the longest side we
must have a < c and b < c, i.e a < 10 and b < 10. Now use a calculator
to approximate the values of a and b (make sure your calculator is in
degree mode).
sin 40◦ =
b
=⇒ b = 10 sin 40◦ ≈ 6.43 cm
10
a
=⇒ a = 10 cos 40◦ ≈ 7.66 cm
10
This seems reasonable since 6.43 < 10 and 7.66 < 10. If either of your
answers were larger than 10 you would know you made a mistake.
cos 40◦ =
3. There are two ways we can draw a picture to set up this problem, as
π
shown below (θ = , for this problem make sure your calculator is in
8
Radian mode).
2
Each picture will give a different answer. If we use the first picture we
have
π
3
3
sin = =⇒ c =
π ≈ 7.84 m
8
c
sin
8
If we use the second picture we have
cos
π
3
3
= =⇒ c =
π ≈ 3.25 m
8
c
cos
8
So there are two possible answers.
4. Let d be the distance from A to C. Then
tan 40◦ =
d
=⇒ d = 100 tan 40◦ ≈ 83.9 ft
100
5. Let h be the height of the Eiffel Tower. Then
tan(85.361◦ ) =
h
=⇒ h = 80 tan(85.361◦ ) ≈ 985.91 ft
80
6. Let d1 be the distance along the ground from Soldier Field to a point
directly beneath the blimp. Let d2 be the distance along the ground
3
from Adler Planetarium to the same point directly beneath the blimp.
If you think about the picture for a minute you will see that
tan(32◦ ) =
500
500
=⇒ d1 =
≈ 800.17 ft
d1
tan(32◦ )
tan(23◦ ) =
500
500
=⇒ d2 =
≈ 1177.93 ft
d2
tan(23◦ )
and
So the distance between Soldier Field and Adler Planetarium is
d1 + d2 =
500
500
+
≈ 1978.09 ft
◦
tan(32 ) tan(23◦ )
7. Let h1 be the total height of the tower. Let h2 be the height of the
tower up tp the top of the Sky Pod. Then
tan 24.4◦ =
h1
=⇒ h1 = 4000 tan 24.4◦
4000
h2
=⇒ h2 = 4000 tan 20.1◦
4000
So the height of the Sky Pod is
tan 20.1◦ =
h1 −h2 = 4000 tan 24.4◦ −4000 tan 20.1◦ = 4000(tan 24.4◦ −tan 20.1◦ ) ≈ 350.69 ft
Note that I did not plug anything into my calculator until the last step
in order to reduce any rounding errors.
8. (a) Let d be the distance between the two buildings. Then
tan 34◦ =
1776
1776
=⇒ d =
≈ 2633 ft
d
tan 34◦
(b) Let h be the height of the Freedom Tower above the office building.
Then
h
tan 20◦ =
=⇒ h = 2633 tan 20◦ ≈ 958 ft
2633
So the height of the office building is 1776 − h ≈ 818 ft
9. Consider the following unit circle:
4
(a) θ = 0 corresponds to the point (1,0).
sin 0 =
0
=0
1
1
=1
1
0
tan 0 = = 0
1
cos 0 =
(b) θ =
π
corresponds to the point (0,1).
2
sin
π
1
= =1
2
1
0
π
= =0
2
1
π
1
tan = is undefined
2
0
cos
(c) θ =
2π
= π corresponds to the point (-1,0).
2
sin π =
0
=0
1
−1
= −1
1
0
tan π =
=0
−1
cos π =
5
(d) θ =
3π
corresponds to the point (0,-1).
2
−1
3π
=
= −1
sin
2
1
3π
0
cos
= =0
2
1
3π
−1
tan
=
is undefined
2
0
For the remaining problems we will refer to the following three diagrams
below and the unit circle above:
6
40π
. First reduce the angle θ.
3
40π
π
π
π
4π
θ=
= 13π + = 12π + π + ≡ π + =
3
3
3
3
3
√
4π
corresponds to the point (−1, − 3) with r = 2. So
and
3
√
2
− 3
csc θ = √
sin θ =
2
− 3
10. Find the six trig functions of θ =
cos θ =
−1
2
√
− 3 √
tan θ =
= 3
−1
−53π
11. Find cos
.
6
sec θ = −2
1
cot θ = √
3
−53π
−5π
−5π
= −8π +
≡
6
6
6
√
−5π
and
corresponds to the point (− 3, −1) with r = 2.
6
√
− 3
.
So cos θ =
2
θ=
7
12. Find tan(−87π).
θ = −87π = −86π − π ≡ −π
=⇒ tan(−87π) = tan(−π) = 0
13. Find sin
26π
.
4
θ=
26π
2π
2π
π
= 6π +
≡
=
4
4
4
2
=⇒ sin θ = sin
14. Find csc
π
=1
2
103π
.
4
θ=
3π
3π
3π
7π
103π
= 25π +
= 24π + π +
≡π+
=
4
4
4
4
4
√
7π
corresponds to the point (1, −1) with r = 2.
4
√
√
r
2
So csc θ = =
=− 2
y
−1
−3π
−3π
15. Find cot
. Note θ =
does not need to be reduced since it
4
4
−3π
corresponds to the point (−1, −1)
is less than a full rotation. Now
4
√
with r = 2.
x
−1
So cot θ = =
=1
y
−1
46π
16. Find sec
.
6
and
θ=
46π
4π
4π
4π
10π
5π
= 7π +
= 6π + π +
≡π+
=
=
6
6
6
6
6
3
√
5π
corresponds to the point (1, − 3) with r = 2.
3
r
2
So sec θ = = = 2
x
1
and
8