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

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Solution

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

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

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

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