Question

In the given $\Delta PQR,\angle Q={90}^o.S$ is a point on $PQ$ such that $\angle SRQ={35}^o{18}^o$ and $\angle PRS=3^o{30}^o$. If $QR-200m$, find the length of $PS$.

Answer 1

$0^{\prime }\to 1^o$
$1^{\prime }\to {\left(\frac{1}{60}\right)}^o$
${18}^{\prime }\to \frac{1}{60}\times 1863=(0.3{)}^o$
${35}^o{18}^{\prime }=(35+0.3)6o$
$\theta =\left(35.3\right)$
$3^o{30}^{\prime }=(3+0.5{)}^o$
$\varphi =(3.5{)}^o$
$\tan \theta =\frac{SQ}{RQ}$
$\tan (\theta +\varphi )=\frac{As}{RQ}$
$\frac{\tan \theta }{\tan (\theta +\varphi )}=\frac{SQ}{PQ}$
$\tan \theta$