Question

Choose the correct answer from the alternatives given :
In $\Delta PQR$, PS is the bisector of $\angle P$ and PT $\perp$ QR then $\angle TPS$ is equal to

• A. $\angle Q+\angle R$
• B. ${90}^{\circ }+1/2\angle Q$
• C. ${90}^{\circ }+1/2\angle R$
• D. $1/2(\angle Q-\angle R)$

Answer 1

The correct option is D $1/2(\angle Q-\angle R)$

We know that,
$\angle TPS=\frac{1}{2}(\angle Q-\angle R)$