Question

In right angled triangle PQR, if $\angle Q={90}^{\circ }$,$PR=5,QR=4$ then find PQ and hence find $\tan R$.

Answer 1

R.E.F.image

In $△PQR,\angle PQR={90}^{\circ }$

BY Pythagoras theorem,

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

$\Rightarrow (5{)}^2=P{Q}^2+(4{)}^2$

$\Rightarrow (5{)}^2-(4{)}^2=P{Q}^2$

$\Rightarrow P{Q}^2=25-16$

$\Rightarrow P{Q}^2=9$

$\Rightarrow PQ=\sqrt{9}$

$\Rightarrow PQ=3$

$\tan R$$=\frac{PQ}{QR}=\frac{3}{4}.$