Question : In $\triangle \mathrm{ABC}, \angle \mathrm{A}=50^{\circ}$, BE and CF are perpendiculars on AC and AB at E and F, respectively. BE and CF intersect at H. The bisectors of $\angle \mathrm{HBC}$ and $\angle H C B$ intersect at P. $\angle B P C$ is equal to:
Option 1: $155^\circ$
Option 2: $100^\circ$
Option 3: $115^\circ$
Option 4: $120^\circ$
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Correct Answer: $155^\circ$
Solution :
Since H is the orthocentre,
$\angle BHC = 180^\circ - 50^\circ = 130^\circ$
Now, P is the incentre of $\triangle BHC$
$\angle BPC = 90^\circ +\frac{1}{2}\angle BHC= 90^\circ +\frac{130^\circ}{2}= 90^\circ +65^\circ= 155^\circ$
Hence, the correct answer is $155^\circ$.
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