Question

P is a point inside $△ABC$ If $\angle PBA={20}^{\circ },\angle BAC={50}^{\circ }and\angle PCA={35}^{\circ }$ then the measure of $\angle BPC$ is

• A. ${65}^{\circ }$
• B. ${75}^{\circ }$
• C. ${90}^{\circ }$
• D. ${105}^{\circ }$

Answer 1

The correct option is D ${105}^{\circ }$

Sum of the three angles of a triangle is ${180}^{\circ }$

Lets assume $\angle PBC=x$ and $\angle PCB=y$

So, $\angle BAC+\angle ABC+\angle BCA={180}^{\circ }$

Now, it is given that $\angle BAC={50}^{\circ }$

$\angle ABC=\angle ABP+\angle PBC={20}^{\circ }+x$

$\angle BCA=\angle BCP+\angle PCA=y+{35}^{\circ }$

So, ${50}^{\circ }+{20}^{\circ }+x+y+{35}^{\circ }={180}^{\circ }$

${105}^{\circ }+x+y={180}^{\circ }$

$x+y={75}^{\circ }$

In triangle $\angle PBC$, sum of all angles is also ${180}^{\circ }$

$x+y+\angle BPC={180}^{\circ }$

${75}^{\circ }+\angle BPC={180}^{\circ }$

$\angle BPC={105}^{\circ }$

Hence the answer is option D

Alternative solution would be to use the theorem that external angle is equal to sum of interior opposite angles.