Trigonometric Ratios (sine, Cosine, and Tangent Ratios) Worksheet

Trigonometric Ratios (sine, Cosine, and Tangent Ratios) Worksheet
• Page 1
1.
What is the value of sin 45o - cos 45o?
 a. 1 b. 2 c. 3

Solution:

From the trigonometric tables sin 45o = 0.7071 and cos 45o = 0.7071

sin 45o - cos 45o = 0.7071 Ã¢â‚¬â€œ 0.7071 = 0
[Substitute the values of sin 45o and cos 45o.]

The value of sin 45o - cos 45o = 0.

2.
Round the value of sin 36o to the nearest hundredth.
 a. 0.59 b. 0.69 c. 0.68 d. 0.81

Solution:

sin 36o = 0.58778525229247312916870595463907
[Find using calculator.]

The value of sin 36o is approximately equal to 0.59.
[Rounded to nearest hundredth.]

3.
sin A =
 a. (Side adjacent to A)/Side opposite to A b. Hypotenuse/Side opposite to A c. (Side opposite to A)/Hypotenuse d. (Side opposite to A)/Side adjacent to A

Solution:

sin A = (Side opposite to A)/(Hypotenuse)

4.
What is the value of tan R in the figure?

 a. $\frac{3}{5}$ b. $\frac{4}{3}$ c. $\frac{3}{4}$ d. $\frac{5}{3}$

Solution:

tan R = (Opposite side of R)/(Adjacent side of R).

Opposite side of R = PQ = 3
adjacent side of RP = 4

In ΔABC, tan R = PQ / RP

tan R = 3 / 4
[Substitute the values of PQ and RP.]

5.
What is the length of side RP in ΔPQR?

 a. 1 foot b. 4 feet c. 6 feet d. 3 feet

Solution:

From the figure, PQ = 6 feet and Q = 45o.

In the figure, RP and PQ are opposite and adjacent sides of angle Q respectively.

tan Q = Opposite side/Adjacent side.

In ΔPQR, tan Q = RP / PQ

tan 45o = RP / 6
[Substitute the values of Q and PQ.]

1 = RP / 6
[Substitute the value of tan 45o from table.]

RP = 6
[Simplify.]

The length of side RP is 6 feet.

6.
What is the length of AC, if the side of each small square is 1 unit?

 a. 6√2 units b. 6 units c. 8√2 units d. 4√2 units

Solution:

sin θ = Opposite side/Hypotenuse.

sin 45o = BC / AC
[From the figure.]

From the figure, BC = 4 units

sin 45o = 4 / AC
[Substitute BC.]

1 / √2 = 4 / AC
[sin 45o = 1 / √2]

AC = 4√2
[Cross multiply.]

So, the length of AC is 4√2 units.

7.
ABC is a right triangle. What is the measure of $\angle$ACB, if AB = 2 cm and AC = 4 cm?

 a. 30o b. 60o c. 45o d. 15o

Solution:

Let ACB = q

In a right triangle, sin q = opposite side / hypotenuse
[Write the formula for the sin ratio.]

= ABAC
[Since the opposite side to ACB is AB and the hypotenuse of the triangle is AC.]

= 24 = 12
[Substitute AB = 2 and AC = 4 and simplify.]

= Sin 30o

So, ACB = 30o.

8.
Find the value of $a$, if ΔABC is a right triangle.

 a. 8.71 b. 9.2 c. 10 d. 6.25

Solution:

sin A = opposite / hypotenuse
[Choose an appropriate trigonometric ratio.]

sin A = BC / AC

sin 35o = 5 / a
[Substitute the values.]

a x sin 35o = 5
[Multiply each side by a.]

a x sin 35o/sin 35o = 5/sin 35o
[Divide each side by sin 35o.]

a = 5/sin 35o
[Simplify.]

a = 5 / 0.5735
[Use table or calculator to find the value of sin 35o.]

a = 8.71
[Divide.]

The value of a is 8.71.

9.
A vertical pole is 70 m high. Find the angle formed by the pole at a point 70 m away from its base.

 a. 30o b. 75o c. 60o d. 45o

Solution:

Let AB be the height of the pole.

The height of the pole, AB = 70 m

Let BC be the distance from the base of the pole to the point where the angle is to be measured. So, BC = 70 m

tan C = opposite side/adjacent side
[Choose an appropriate trigonometric ratio.]

From ΔABC, tan C = AB / BC

tan C = 7070 = 1
[Substitute and simplify.]

From the trigonometric tables, tan 45o = 1

So, the angle formed by the pole at the point 70 m away from its base is 45o.
[As tan C = 1 and tan 45o = 1, C = 45o.]

10.
What is the value of sin A in the figure?

 a. $\frac{12}{13}$ b. $\frac{5}{13}$ c. $\frac{13}{5}$ d. $\frac{13}{12}$

Solution:

sin A = (Side opposite to A)/Hypotenuse

In the triangle, BC is the side opposite to A and AC is the hypotenuse.

sin A = BCAC

= 513
[Substitute BC = 5 and AC = 13.]