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

**Solution:**

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

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 15 cot A = 8

⇒ AB/BC = 8/15

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

Applying 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.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 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

**Summary:**

If 15 cot A = 8, the value of sin A = 15/17 and sec A = 17/8

**☛ Related Questions:**

- Given sec θ = 13/12, calculate all other trigonometric ratios.
- If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
- If cotθ = 7/8, evaluate: (i) (1 + sinθ)(1 - sinθ) / (1 + cosθ)(1 - cosθ), (ii) cot2θ
- If 3 cot A = 4, check whether (1 - tan2 A) / (1 + tan2 A) = cos2 A - sin2 A or not.

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