$\triangle PQR$માં $\angle Q$ કાટખૂણો છે અને $PR + QR = 25$ સેમી  અને $PQ = 5$ સેમી હોય, તો $\sin P, \cos P$ અને $\tan$ $P$ શોધો.

Given that, $PR + QR =25$

$PQ =5$

Let $PR$ be $x$.

Therefore, $QR =25-x$

Applying Pythagoras theorem in $\triangle PQR$, we obtain

$PR ^{2}= PQ ^{2}+ QR ^{2}$

$x^{2}=(5)^{2}+(25-x)^{2}$

$x^{2}=25+625+x^{2}-50 x$

$50 x=650$

$x=13$

Therefore, $PR =13 \,cm$

$Q R=(25-13) \,cm =12\, cm$

$\sin P =\frac{\text { Side opposite to } \angle P }{\text { Hypotenuse }}=\frac{ QR }{ PR }=\frac{12}{13}$

$\cos P =\frac{\text { Side adjacent to } \angle P }{\text { Hypotenuse }}=\frac{ PQ }{ PR }=\frac{5}{13}$

$\tan P =\frac{\text { Side opposite to } \angle P }{\text { Side adjacent to } \angle P }=\frac{ QR }{ PQ }=\frac{12}{5}$