CIRCLES (FINDING ANGLES)

Question

In the given figure a semi-circle is drawn with centre M and AB is diameter. If $\measuredangle MQN= \measuredangle MPN= 10^\text{o}$ and $\ \measuredangle AMQ=40^\text{o}$, then the measure of $\angle PMB$ equals

$$(1)\quad20^\text{o}\quad\quad\quad\quad(2)\quad30^\text{o}$$

My attempts:

Answer 1

Let $\measuredangle PMN=x$.

Since $\measuredangle MQN=\measuredangle MPN$, we obtain that $MQPN$ is cyclic.

Thus, $$\measuredangle NQP=\measuredangle NMP=x,$$ which says $$\measuredangle QPN=10^{\circ}+x+10^{\circ}=20^{\circ}+x.$$ Id est, $$20^{\circ}+x=40^{\circ},$$ which gives $$x=20^{\circ}.$$

Answer 2

I think PMB is 20°. Here is my explanation: The quadrilateral MNPQ has to be cyclic because angle MQN=angle MPN. So angle QMN +angle NPQ =180. Therefore, angle QPN= 40°. So, angle QPM= 40-10=30°. Now,angle QMA=40°. SO angle QPA=20°. (Half of QMA).So angle APM=10°. now angle APB=90° So angle NPB=70°. now in traingle MPB, angle MPB=MBP. so angle PMB=20°