In the given figure, find the measure of ∠RQT, if PQ = QR and ∠QPR = 60°. 

$∆PQR$ is isosceles since $PQ=QR$.

$\therefore \angle QPR=\angle QRP={60}^{°}$

$\angle RQT$ is the exterior angle of $∆PQR$ which is equal to the sum of interior opposite angles $\angle P$ and $\angle R.$

Hence, $\angle RQT={60}^{° }+ {60}^{°}={120}^{°}.$