# NCERT Maths Class 10 Chapter 8, Exercise 8.1 | Question 3- If sin A = 3/4, Calculate cos A and tan A

Let us assume a right-angled triangle ABC, right-angled at B

Let us assume a right-angled triangle ABC, right-angled at B

Given: Sin A = 3/4

We know that the Sin function is equal to the ratio of the length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle and we get, AC2=AB2 + BC2

Substitute the value of AC and BC

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

16k2−9k2 =AB2

AB2=7k2

Therefore, AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7

**Therefore, tan A = 3/√7**