Question

3. If sin A = mit calculate cos A and tan A.​

Answer 1

Given that:

sin A = 3/4

⇒ BC/AC = 3/4

Let BC be 3k. Therefore, hypotenuse AC will be 4k where k is a positive integer.

Applying Pythagoras theorem on ∆ABC, we obtain:

AC2 = AB2 + BC2

AB2 = AC2 - BC2

AB2 = (4k)2 - (3k)2

AB2 = 16k2 - 9k2

AB2 = 7 k2

AB = √7 k

cos A = side adjacent to ∠A / hypotenuse = AB/AC = √7 k / 4k = √7/4

tan A = side opposite to ∠A / side adjacent to ∠A = BC/AB = 3k / √7 k = 3/√7

Thus, cos A= √7/4 and tan A = 3/√7

Hope it will help you..

Answer 2

Given :

The value of Sin A = 3/4

Step 1:

Use the square relation ,



Therefore ,

cos 2 (A) = 1 - sin2 (2)

cos(A) = 1-9/6



cos(A)=✓7/16

Step 2 :

Use the fundamental trigonometric relation.

tan(0)= SIN(0)/COS(0)

Therefore

TAN (A)=(3/4 X 4/✓7)

TAN (A) = 3/✓7

COS(A) = ✓7/4

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