Question

In the given figure, $PT\perp QP$ and PS bisects $\angle QPR.$ If $\angle Q={60}^{o}and\angle R={30}^o$, find $\angle TPS$.

Answer 1

$\angle PQR+\angle QRP+\angle RPQ={180}^o$ (sum of all $\angle$'s of $△$)

$60+30+\angle RPCS={180}^o$

$\angle RPQ=180-90$

$\angle RPQ={90}^o$

According to question

$PS$ is the bisector of $\angle P$

$\therefore PQS=\angle SPR={45}^o(\angle P=90)$

Now In $\Delta PTS$

$\angle PTS+\angle TSP+\angle SPT={180}^o$ (sum of all $\angle$'s of $△$)

${90}^o+75+\angle TPS=180$

$\angle TPS=180-165$

$\angle TPS=15$

$\therefore$ Thus, $\angle TPS={15}^o$