A) \[{{60}^{o}}\]
B) \[{{140}^{o}}\]
C) \[{{120}^{o}}\]
D) \[{{100}^{o}}\]
Correct Answer: C
Solution :
\[\Delta PQR\] is isosceles since \[PQ=QR\]. \[\therefore \] \[\angle QPR=\angle QRP={{60}^{o}}\] \[\angle RQT\] is the exterior angle of \[\Delta PQR\] which is equal to the sum of interior opposite angles \[\angle P\] and \[\angle R\]. Hence, \[\angle RQT={{60}^{o}}+{{60}^{o}}={{120}^{o}}\].You need to login to perform this action.
You will be redirected in
3 sec