# NCERT Maths Class 10 Chapter 8, Exercise 8.1 | Question 4- Given 15 cot A = 8, find sin A and sec A

Solution for "Given 15 cot A = 8, find sin A and sec A"

**Solution for "Given 15 cot A = 8, find sin A and sec A"**

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

Given: 15 cot A = 8

So, Cot A = 8/15

We know that the Cot function is equal to the ratio of the length of the adjacent side to the opposite side.

Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15

Let AB be 8k and BC will be 15k

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 AB and BC

AC2= (8k)2 + (15k)2

AC2= 64k2 + 225k2

AC2= 289k2

**Therefore, AC = 17k**

Now, we have to find the value of sin A and sec A

We know that Sin (A) = Opposite side /Hypotenuse

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

Sin A = BC/AC = 15k/17k = 15/17

**Therefore, sin A = 15/17**

Since secant or sec function is the reciprocal of the Cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.

Sec (A) = Hypotenuse/Adjacent side

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

AC/AB = 17k/8k = 17/8

**Therefore sec (A) = 17/8**