Question

In the $\Delta$ PQR, PS is the bisector of $\angle P$ and PT $\perp$ QR, then $\angle TPS$ is equal to

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

Answer 1

The correct option is D $\frac{1}{2}(\angle Q-\angle R)$

PS is the bisector of $\angle QPR$
$\therefore \angle 1+\angle 2=\angle 3[Eq.(1\left)\right]\Rightarrow \angle Q={90}^{\circ }-\angle 1\angle R={90}^{\circ }-\angle 2-\angle 3So,\angle Q-\angle R=[{90}^{\circ }-\angle 1]-[{90}^{\circ }-\angle 2-\angle 3]\Rightarrow \angle Q-\angle R=\angle 2+\angle 3-\angle 1=\angle 2+(\angle 1+\angle 2)-\angle 1[FromEq.(1\left)\right]$

$\Rightarrow \angle Q-\angle R=2\angle 2\Rightarrow \frac{1}{2}(\angle Q-\angle R)=\angle TPS$