# If sin A = 3/4, calculate cos A and tan A.

**Solution:**

We will use the basic formula of sine, cosine, and tangent functions to solve the question.

Let's draw a figure according to the given question.

Let ∆ABC be a right-angled triangle, right-angled at point B.

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:

AC^{2} = AB^{2} + BC^{2}

AB^{2} = AC^{2} - BC^{2}

AB^{2 }= (4k)^{2} - (3k)^{2}

AB^{2} = 16k^{2} - 9k^{2}

AB^{2} = 7 k^{2}

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

**Video Solution:**

## If sin A = 3/4 calculate cos A and tan A.

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.1 Question 3:

**Summary:**

If sin A = 3/4, the value of cos A= √7/4 and tan A = 3/√7.