In triangle $P Q R$ , right-angled at $Q$ , $P R + Q R = 25 c m$ and $P Q = 5 c m .$ The value of $tan P$ is $\frac{12}{q}$ , $q$ is

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Solution

$I n t h e △ P Q R, {P R}^{2} = {P Q}^{2} + {Q R}^{2} x^{2} = 5^{2} + {(25 - x)}^{2} x^{2} {- (25 - x)}^{2} = 25 2 x - 25 = 1 x = 13 c m ∴ tan (P) = \frac{Q R}{Q P} = \frac{25 - 13}{5} = \frac{12}{5}$

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