If 15 cot A = 8, find sin A and sec A. $[2 Marks]$

Open in App

Solution

Let us draw a right triangle ABC, right angled at B.
It is given that: 15 cot A = 8
$c o t A = \frac{8}{15} = \frac{A B}{B C}$

$ \sum $

Let, AB = 8K, BC = 15K
Using Pythagoras theorem, we have
$A C^{2} = A B^{2} + B C^{2} = (8 K)^{2} + (15 K)^{2} = 64 K^{2} + 225 K^{2} = 289 K^{2} ∴ A C = 17 K [1 Mark]$

$∴ s i n A = \frac{B C}{A C} = \frac{15 K}{17 K} = \frac{15}{17}$ , and
$s e c A = \frac{A C}{A B} = \frac{17 K}{8 K} = \frac{17}{8}$ $[1 Mark]$

Suggest Corrections

Similar questions

If 15 cot A = 8, find the values of sin A and sec A.

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

Given 15 cot A = 8. Find sin A and sec A

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

15 cot A = 8

sin A

sec A .

Join BYJU'S Learning Program