Question

In a triangle ABC, $\angle A={60}^{\circ }$ and the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively.

• A. $\angle QOR={120}^{\circ }$
• B. $\angle QPR={60}^{\circ }$
• C. $\angle QOR={90}^{\circ }$
• D. $\angle QPR={50}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A

$\angle QOR={120}^{\circ }$

B

$\angle QPR={60}^{\circ }$

The incircle touches the sides of the triangle ABC.
and $OP\perp BC,OQ\perp AC.OR\perp AB$
In quad. AROQ

$\angle ORA={90}^{\circ },\angle OQA={90}^{\circ }and\angle A={60}^{\circ }$

$\angle QOR={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{60}^{\circ })$ = ${120}^{\circ }$

Now are RQ subtends $\angle QOR$ at the centre and $\angle QPR$ at the remaining part of the circle.

$\therefore \angle QPR=\frac{1}{2}\angle QOR$

$=\frac{1}{2}\times {120}^{\circ }={60}^{\circ }$