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# Given 15 cot A = 8, find sin A and sec A. Q.4

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How to solve the tricky question of introduction to trigonometry of ncert class 10 , find the solution of exercise 8.1 question no.4, please help me to find the best way to solve this question of trigonometry .Given 15 cot A = 8, find sin A and sec A.

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1. 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, cot function is the equal to the ratio of 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.