Question

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

Answer 1

We have a right-angled $△PQR$ in which $\angle Q={90}^o$

Let $QR=xcm$

Therefore, $PR=(25-x)cm$

By Phythagoras Theorem, we have

$P{R}^2=P{Q}^2+Q{R}^2$

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

$\Rightarrow (25-x-x)(25-x+x)=25$

$\Rightarrow (25-2x)25=25$ $\Rightarrow 25-2x=1$

$\Rightarrow 25-1=2x$ $\Rightarrow 24=2x$

$\therefore x=12cm$

Hence, $QR=12cm$

$PR(25-x)cm=25-12=13cm$

$PQ=5cm$

$\therefore \sin P=\frac{QR}{PR}=\frac{12}{13},\cos P=\frac{PQ}{PR}=\frac{5}{13};\tan P=\frac{QR}{PQ}=\frac{12}{5}$