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

Solution :

We have,triangle

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 = \(AC\over AB\) = \(!7k\over 8k\) = \(17\over 8\)

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