Practice exercises - Exam 1 (Sections 1.1 through 2.4)
1. Suppose the angle α has measure 14◦ .
(a) What is the measure of the complement of α?
(b) In standard position, the supplement of α lies in what quadrant?
(c) Give an example of another angle coterminal with α.
2. Suppose the point (−2, 5) lies on the terminal side of θ. Compute sin θ and cot θ.
3. Find the distance between the points (1, −2) and (4, 3).
4. Assuming cos θ =
5. If sin θ =
√1
5
1
4
and θ lies in the fourth quadrant, find the values of sin θ and sec θ.
, what can we say about the quadrant where the terminal side of θ lies?
6. Show that each of the following statements are identities by transforming the left hand
side into the right hand side:
(a)
cos θ
sec θ
= cos2 θ
(b) cos θ (csc θ + tan θ) = cot θ + sin θ
(c) sin θ (sin θ + cos θ cot θ) = 1
7. Find the exact values of the following expressions:
(b) sin 60◦ + cos 30◦
(a) tan2 45◦
8. Approximate the following values (include 4 digits after the decimal point):
(a) csc 12◦
(b) the angle θ with cot θ = 4
9. In right triangle ABC with right angle C, suppose b = 2 feet and c = 3 feet.
(a) Find the length a (to the nearest hundredth).
(b) Find the measure of angle A (to the nearest tenth).
10. In the diagram below, suppose A = 23◦ , D = 40◦ , and the length CD = 5 inches. Find the
length AB (to the nearest hundredth).
B
A
C
D
11. If the angle of elevation to the sun is 62◦ when a building casts a shadow 37 feet long,
what is the height of the building (to the nearest hundredth)?
12. A ship’s intended course is due north. However, the ship actually travels at a bearing of N
2.8◦ E. If the ship is traveling at 14 miles per hour, how many miles off its intended course
will it be after 2 hours (to the nearest hundredth)?
Answers
1. (a) 76◦
(b) 166◦ lies in the second quadrant
(c) coterminal angles take the form 14 plus or minus a multiple of 360; examples include
374◦ and −346◦ .
2. sin θ =
y
r
=
√5
29
and cot θ =
x
y
= − 25
3. Using the distance formula we get
√
32 + 52 =
√
34
4. In the fourth quadrant, sin θ is negative.
From sin θ + cos θ = 1, the value cos θ =
Using a reciprocal identity, sec θ = 4.
2
2
1
4
!
gives us sin θ = − 1 −
" #2
1
4
=−
√
15
4
5. θ must be in the first or second quadrant to make the sine positive.
6. Your answers should include a sequence of steps that someone can follow:
(a)
cos θ
sec θ
=
cos θ/1
1/cos θ
=
cos θ
1
·
cos θ
1
= cos2 θ
(b) cos θ (csc θ + tan θ) = cos θ ·
1
sin θ
+ cos θ ·
sin θ
cos θ
(c) sin θ (sin θ + cos θ cot θ) = sin θ · sin θ + sin θ ·
= sin2 θ + cos2 θ = 1
cos θ
θ sin θ
+ coscos
= cot θ + sin θ
sin θ
θ
θ
θ cos θ
= sin2 θ + sin θ cos
cos θ cos
sin θ
sin θ
=
2
7. (a) (1)
=√1
√
√
3
(b) 2 + 23 = 3
8. (a) 1/(sin 12◦ ) ≈ 4.8097
(b) θ = tan−1 (1/4) ≈ 14.0362◦
9. (a) Using the "Pythagorean
Theorem, a =
#
−1 2
◦
≈ 48.2
(b) A = cos
3
√
32 − 22 =
√
5 ≈ 2.24 feet
◦
tan 40
10. AB = 5sin
≈ 10.74 inches.
23◦
To find this, from the right triangle BCD, tan 40◦ = BC
gives BC = 5 tan 40◦ ≈ 4.1955
5
40◦
Using this length in the right triangle ABC, sin 23◦ = BC
= 5 tan
. Solve this for AB.
AB
AB
11. The height is 37 tan 62◦ ≈ 69.59 feet.
The 37 foot shadow along the ground is adjacent to the 62◦ angle.
12. The distance off course is 28 sin 2.8◦ ≈ 1.37 miles
The ship travels 28 miles in two hours; this forms the hypotenuse and we are interested
in the side opposite 2.8◦ .