Question

In triangle ABC, the bisectors of angles B and C meet at P.
Then , $\angle BPC={60}^o+\frac{1}{2}\angle A$.
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△BPC$,
Sum of angles = 180
$\angle BPC+\angle PBC+\angle PCB=180$
$\angle BPC=180-(\angle PBC+\angle PCB)$
$\angle BPC=180-\frac{1}{2}(\angle B+\angle C)$ (PB and PC bisects $\angle B$ and $\angle C$ respectively)
$\angle BPC=180-\frac{1}{2}(180-\angle A)$ (Consider sum of angles of triangle ABC)
$\angle BPC=180-90+\frac{1}{2}\angle A$
$\angle BPC=90+\frac{1}{2}\angle A$