question_answer

BC is the diameter of a semicircle. The sides AB and AC of a triangle ABC meet the semicircle at P and Q respectively. PQ subtends \[{{140}^{o}}\]at the centre of the semicircle. Then \[\angle A\] is _____.

A) \[{{10}^{o}}\]             

B)                    \[{{20}^{\text{o}}}\]          

C)        \[{{30}^{o}}\]            

D)        \[{{40}^{o}}\]

Correct Answer: B

Solution :

\[PO=OQ=BO=OC=r\] \[{{140}^{o}}+26={{180}^{o}}\][linear pair] \[\therefore \]    \[\theta ={{20}^{o}}\] Now, in \[\Delta PBO\] \[2\alpha +\theta ={{180}^{o}}\]   \[\therefore \]  \[\alpha ={{80}^{o}}\] \[\angle APO={{100}^{o}}\] (Exterior angle)      Similarly,  \[\angle AQO={{100}^{o}}\]  In quadrilateral APOQ, \[{{100}^{o}}+{{100}^{o}}+{{140}^{o}}+\angle PAQ={{360}^{o}}\] \[\therefore \]   \[\angle PAQ={{20}^{o}}\]