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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

enter image description here

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

My attempts: enter image description here

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  • $\begingroup$ shall i post it in answers? $\endgroup$ – all about everything Mar 24 at 6:31
  • $\begingroup$ Post your trying in your starting post. $\endgroup$ – Michael Rozenberg Mar 24 at 6:36
  • $\begingroup$ Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here! $\endgroup$ – BadAtGeometry Mar 24 at 6:37
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    $\begingroup$ i posted my attempts and thank you very much for asking my attempts $\endgroup$ – all about everything Mar 24 at 6:42
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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 $$x=20^{\circ}.$$

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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°enter image description here

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