Question

Given: In Angle ABC, angleB = 90' If 15 cot A=8,
find sin A and see A.​

Answer 1

Answer:

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Answer 2

Answer:

Sin A = 15/17

Sec A=17/8

Explanation:

Let Triangle ABC is a right-angled triangle and angle B is 90°

We know that cot A = B/P

= 8/15

Let AB be 8k and BC be 15k, where K is a positive real number.

By Pythagoras Theoram,

AC² = AB² + BC²

AC²= (8k)² + (15k)²

AC²= 64k² + 225k²

AC²= 289k²

AC²= 17k

Now,

SinA = P/H

= 15k/17k

= 15/17

Sec A = H/B

= 17k/8k

= 17/8

Hope you you all like the solution Thankss!

:-) Waise Question me galti hai last me jahaan SecA ko seeA likh diya hai

Rishabh Kumar