Question

In $\Delta$ABC, m$\angle$BAC$={40}^o$, 'P' is the centre of the circumcircle of $\Delta$ABC. Find m$\angle$PBC.

• A. ${40}^o$
• B. ${50}^o$
• C. ${80}^o$
• D. ${100}^o$

Answer 1

The correct option is B ${50}^o$

Let 'P' be the centre of circumcircle of $△ABC$.
The distance of P from the vertices of triangle will be same i.e $PA=PB=PC$

AS $PB=PC$, then $△PBC$ is an isosceles triangle

Also, $\angle PBC=\angle PCB$

$\therefore \angle B+\angle P+\angle C={180}^o$

But $\angle BPC=2\times \angle BAC={80}^o$

$\therefore \angle B+\angle C+{80}^o={180}^o$

$⟹\angle PBC+\angle PCB=180-80={100}^o$

$⟹2\times \angle PBC={100}^o$ ..... since ($\angle PBC=\angle PCB$)

$\therefore \angle PBC={50}^o=\angle PCB$

Hence, option B is correct.