In $\triangle PQR ,$ right $-$ angled at $Q , PR + QR =25\, cm$ and $PQ =5\, cm .$ Determine the values of $\sin P, \cos P$ and $\tan P$.
In $\triangle PQR ,$ right $-$ angled at $Q , PR + QR =25\, cm$ and $PQ =5\, cm .$ Determine the values of $\sin P, \cos P$ and $\tan P$.
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, $PR =13 \,cm$
$Q R=(25-13) \,cm =12\, cm$
$\sin P =\frac{\text { Side opposite to } \angle P }{\text { Hypotenuse }}=\frac{ QR }{ PR }=\frac{12}{13}$
$\cos P =\frac{\text { Side adjacent to } \angle P }{\text { Hypotenuse }}=\frac{ PQ }{ PR }=\frac{5}{13}$
$\tan P =\frac{\text { Side opposite to } \angle P }{\text { Side adjacent to } \angle P }=\frac{ QR }{ PQ }=\frac{12}{5}$
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