# Given 15 cot A = 8, find sin A and sec A

**Solution:**

We use the basic formula of trigonometric ratios cot, sin and tan to solve the question.

Using cot A, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Let us consider a right-angled

ΔABC, right-angled at B.

cot A= side adjacent to ∠A/side opposite to ∠A = AB/BC

It is given that cot A/15 = 8 ⇒ AB/BC = 8/15

Let AB be 8k. Therefore, BC will be 15 k where k is a positive integer.

Apply Pythagoras theorem in ΔABC, we obtain.

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

AC^{2} =(8k)^{2} + (15k)^{2}

AC^{2} = 64k^{2} + 225k^{2}

AC^{2} = 289k^{2}

AC = 17k

sin A = side opposite to ∠A/hypotenuse = BC/AC = 15k / 17k = 15/17

sec A = hypotenuse/side adjacent to ∠A = AC/AB = 17k / 8k = 17/8

Thus, sin A = 15/17 and sec A = 17/8

**Video Solution:**

## Given 15 cot A = 8, find sin A and sec A

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.1 Question 4:

Given 15 cot A = 8, find sin A and sec A

If 15 cot A = 8, Thus, sin A = 15/17 and sec A = 17/8