GATE 2026 EE GA Q.7: solution and answer

Official answer: BSource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. Draw chord $RQ$ (it is not in the printed figure). $PQ$ is a diameter, and an angle in a semicircle is a right angle, so $\angle PRQ=90^{\circ }$.

2. $T$ lies on line $PR$, beyond $R$. Angles on a straight line at $R$ add to $180^{\circ }$, so $\angle TRQ=180^{\circ }-90^{\circ }=90^{\circ }$.

3. An inscribed angle is half the central angle standing on the same arc. $\angle RQS$ and $\angle ROS$ both stand on the minor arc $RS$, which does not contain $Q$. So $\angle RQS=\frac{80^{\circ }}{2}=40^{\circ }$.

4. $S$ lies on segment $QT$, so $\angle RQT=\angle RQS=40^{\circ }$.

5. The angles of triangle $RQT$ add to $180^{\circ }$: $$\angle RTQ=180^{\circ }-90^{\circ }-40^{\circ }=50^{\circ }$$

6. $S$ lies on $TQ$, so $\angle RTS$ is the same angle as $\angle RTQ$. Therefore $\angle RTS=50^{\circ }$, which is option (B).

7. Check: when two secants from an outside point meet a circle, the angle between them is half the difference of the two arcs they cut off. The far arc is $PQ$, the lower semicircle ($180^{\circ }$). The near arc is $RS$ ($80^{\circ }$). So $\angle RTS=\frac{180^{\circ }-80^{\circ }}{2}=50^{\circ }$.