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: $81^\circ$
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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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Question : $\angle A, \angle B$ and $\angle C$ are three angles of a triangle. If $\angle A- \angle B=15^{\circ}, \angle B - \angle C=30^{\circ}$, then $\angle A, \angle B$ and $\angle C$ are:
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Question : In $\triangle A B C, \angle B=78^{\circ}, A D$ is a bisector of $\angle A$ meeting BC at D, and $A E \perp B C$ at $E$. If $\angle D A E=24^{\circ}$, then the measure of $\angle A C B$ is:
Question : In a triangle ${ABC}, {D}$ is a point on ${BC}$ such that $\frac{A B}{A C}=\frac{B D}{D C}$. If $\angle B=68^{\circ}$ and $\angle C=52^{\circ}$, then measure of $\angle B A D$ is equal to:
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