# BASIC TRIGONOMETRIC RATIOS WORKSHEET

Problem 1 :

In the right triangle PQR given below, find the basic trigonometric ratios of angle θ. Problem 2 :

From the figure given below, find the basic trigonometric ratios of angle θ. Problem 3 :

If sin θ  =  13/85 and cos θ  =  84/85, then find the values of tan θ and cos θ.

Problem 4 :

In triangle ABC, right angled at B, 15sin A = 12. Find the basic trigonometric ratios of angle A and angle C. Problem 1 :

In the right triangle PQR given below, find the basic trigonometric ratios of the angle θ. Solution :

In the triangle shown above, for the angle θ,

opposite side  =  5

hypotenuse  =  13

Then, the basic trigonometric ratios of the angle θ are

 sin θ  =  5/13cos θ  =  12/13tan θ  =  5/12 csc θ  =  13/5sec θ  =  13/12cot θ  =  5/12

Problem 2 :

From the figure given below, find the six trigonometric ratios of the angle θ. In the triangle shown above, by Pythagorean Theorem,

AB2  =  BC2 + CA2

AB2  =  72 + 242

AB2  =  49 + 576

AB2  =  625

AB2  =  252

AB  =  25

In the triangle shown above, for the angle θ,

opposite side  =  7

hypotenuse  =  25

Then, the basic trigonometric ratios of the angle θ are

 sin θ  =  7/25cos θ  =  24/25tan θ  =  7/24 csc θ  =  25/7sec θ  =  25/24cot θ  =  24/7

Problem 3 :

If sin θ  =  13/85 and cos θ  =  84/85, then find the values of tan θ and cos θ.

Solution :

Finding the value of tan θ :

tan θ  =  sin θ / cos θ

tan θ  =  (13/85) / (84/85)

tan θ  =  (13/85)  (85/84)

tan θ  =  (13 ⋅ 85) / (85 ⋅ 84)

tan θ  =  13/84

Finding the value of cot θ :

cot θ  =  84/13

Problem 4 :

In triangle ABC, right angled at B, 15sin A = 12. Find the basic trigonometric ratios of angle A and angle C.

Solution :

Given : 15sin A  =  12

Then,

15sin A = 12

Divide each side by 15.

sin A  =  12/15 -----(1)

We know that,

sin A  =  opposite side / hypotenuse -----(2)

Comparing (1) and (2), we get

opposite side  =  12

hypotenuse  =  15

Let us consider the triangle ABC where right angled at B, with BC  =  12 and AC  =  15. In the triangle shown above, by Pythagorean Theorem,

AC2  =  AB2 + BC2

152  =  AB2 + 122

225  =  AB2 + 144

Subtract 144 from each side.

81  =  AB2 + 144

92  =  AB2

9  =  AB

In the triangle shown above, for the angle A,

opposite side  =  12

hypotenuse  =  15

Then, the basic trigonometric ratios of the angle A are

 sin A  =  12/15  =  4/5cos A  =  9/15  =  3/5tan A  =  12/9  =  4/3 csc A  =  5/4sec A  =  5/3cot A  =  3/4

In the triangle shown above, for the angle C,

opposite side  =  9

hypotenuse  =  15

Then, the basic trigonometric ratios of the angle C are

 sin C  =  9/15  =  3/5cos C  =  12/15  =  4/5tan C  =  9/12  =  3/4 csc C  =  5/3sec C  =  5/4cot C  =  4/3 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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