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

## Solution :

We have,

15 cot A = 8  $$\implies$$  cot A = $$8\over 15$$

Draw a right triangle ABC,

cot A = $$AB\over BC$$ = $$8\over 15$$

If BC = 15k, then AB = 8k, where k is any positive number.

By using Pythagoras Theorem,

$${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$

= $$64k^2$$ + $$225k^2$$ = $$289k^2$$

$$\implies$$  AC = 17k

Thus, sin  A = $$BC\over AC$$ = $$15k\over 17k$$ = $$15\over 17$$

sec A = $$AC\over AB$$ = $$!7k\over 8k$$ = $$17\over 8$$