question_answer

In \[\Delta \,PQR\], if O is the orthocentre and \[\angle \,QOR\,=\,2\,\angle P\] , then \[\angle \,QOR\] is equal to

A) \[90{}^\circ \]

B) \[120{}^\circ \]  

C) \[150{}^\circ \]

D) \[160{}^\circ \]

Correct Answer: B

Solution :

  Let PX, QY, RZ be the altitudes from P, Q, R which meet the opposite sides at X, Y, Z respectively go that their point of intersection O is the orthocentre.                                 From right angled As PQ X and PRX                        \[P{{Q}^{2}}=Q{{X}^{2}}+P{{X}^{2}}\] and        \[P{{R}^{2}}=P{{X}^{2}}+X{{R}^{2}}\] \[\therefore \]  \[PQ=PR\] Similarly \[QR=PR\] Thus    \[PQ=PR=QR\] so that \[\Delta PQR\] is an equilateral \[\therefore \]  \[\angle P={{60}^{o}}\] Hence,  \[\angle QOR=2\angle P\]  (given)                 \[=2\times {{60}^{o}}={{120}^{o}}\]