Question : In $\triangle A B C, \angle A=66^{\circ}$ and $\angle B=50^{\circ}$. If the bisectors of $\angle B$ and $\angle C$ meet at P, then, $\angle B P C-\angle P C A=$?
Option 1: $93^\circ$
Option 2: $91^\circ$
Option 3: $83^\circ$
Option 4: $81^\circ$
Correct Answer: $91^\circ$
Solution :
We know that the sum of all the angles of the triangle is $180^\circ$
⇒ $\angle A + \angle B + \angle C = 180^\circ$
⇒ $66^\circ + 50^\circ + \angle C = 180^\circ$
⇒ $116^\circ + \angle C = 180^\circ$
⇒ $\angle C = (180^\circ -116^\circ)$
⇒ $\angle C = 64^\circ$
Now,
By the angle bisector theorem,
⇒ $\angle BPC = 90^\circ + \frac{\angle A}{2}$
⇒ $\angle BPC = 90^\circ + \frac{66}{2}$
⇒ $\angle BPC = 90^\circ + 33^\circ$
⇒ $\angle BPC = 123^\circ$
Again,
$\angle PCA = \angle \frac{C}{2}= \frac{64}{2}=32^\circ$
Now,
The value of $\angle BPC - \angle PCA$
= $123^\circ - 32^\circ$
= $91^\circ$
Hence, the correct answer is $91^\circ$.
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