# TRIGONOMETRIC RATIOS WORKSHEET

Problem 1 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. Problem 2 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. Problem 3 :

In triangle ABC, right angled at B, 15 sin A = 12. Find the other five trigonometric ratios of the angle A. Also find the six ratios of the angle C. Problem 4 :

In the right triangle PQR shown below, find the value of sin θ and cos θ. Using them, find the value of tan θ and cot θ. Problem 5 :

From the figure given below, find the value of sin θ and cosθ. Using them, find the value of tan θ and cot θ.  Problem 1 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. Solution :

From the triangle shown above,

opposite side  =  5

hypotenuse  =  13

Therefore, Problem 2 :

In the right triangle shown below, find the six trigonometric ratios of the angle θ. Solution :

From the right triangle shown above,

AC  =  24

BC  =  7

By Pythagorean Theorem,

AB2  =  BC2 + CA2

AB2  =  72 + 242

AB2  =  49 + 576

AB2  =  625

AB2  =  252

AB  =  25

Now, we can use the three sides to find the six trigonometric ratios of angle θ.

Therefore, Problem 3 :

In triangle ABC, right angled at B, 15 sin A = 12. Find the other five trigonometric ratios of the angle A. Also find the six ratios of the angle C.

Solution :

Given :  15sin A  =  12.

Then,

sin A  =  12/15

Therefore,

opposite side  =  12

hypotenuse  =  15

Let us consider the right triangle ABC where right angled at B, with

BC  =  12

AC  =  15 By Pythagorean theorem,

AC2  =  AB2 + BC2

152  =  AB2 + 122

225  =  AB2 - 144

225 - 144  =  AB2

81  =  AB2

9=  AB2

9  =  AB

Now, we can use the three sides to find the five trigonometric ratios of angle A and six trigonometric ratios of angle C.

Therefore, Problem 4 :

In the right triangle PQR shown below, find the value of sin θ and cos θ. Using them, find the value of tan θ and cot θ. Solution :

From the right triangle shown above,

opposite side  =  5

hypotenuse  =  13

Therefore,

sin θ  =  PQ/RQ  =  5/13

cos θ  =  PR/RQ  =  12/13

tan θ  =  sin θ / cos θ  =  (5/13) ÷ (12/13)

tan θ  =  (5/13)  (13/12)

tan θ  =  5/12

cot θ  =  cos θ / sin θ  =  (12/13) ÷ (5/13)

cot θ  =  (12/13)  (13/5)

cot θ  =  12/5

Problem 5 :

From the figure given below, find the value of sin θ and cos θ. Using them, find the value of tan θ and cot θ. Solution :

From the figure given above, AC = 24 and BC = 7.

By Pythagorean theorem,

AB2  =  BC2 + CA2

AB2  =  72 + 242

AB2  =  49 + 576

AB²  =  49 + 576

AB2  =  625

AB2  =  252

AB  =  25

Now, we can use the three sides to find the six trigonometric ratios of angle θ.

Therefore,

opposite side  =  7

hypotenuse  =  25

Therefore,

sin θ  =  BC/AB  =  7/25

cos θ  =  AC/AB  =  24/25

tan θ  =  sin θ / cos θ  =  (7/25) ÷ (24/25)

tan θ  =  (7/25)  (25/24)

tanθ  =  7/24

sin θ  =  BC/AB  =  7/25

cos θ  =  AC/AB  =  24/25

cot θ  =  cos θ / sin θ  =  (24/25) ÷ (7/25)

cot θ  =  (24/25)  (25/7) Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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