In $\triangle PQR ,$ right $-$ angled at $Q , PR + QR =25\, cm$ and $PQ =5\, cm .$ Determine the values of $\sin P, \cos P$ and $\tan P$.

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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, $PR =13 \,cm$

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

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

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

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

1043-s15

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