Question

In the given figure, PQ is a diameter of the circle whose centre is O. Given $\angle ROS={42}^{\circ }$, $\angle RTS$ = ____________

• A. $={69}^{\circ }$
• B. $={29}^{\circ }$
• C. $={89}^{\circ }$
• D. $={19}^{\circ }$

Answer 1

The correct option is A

$={69}^{\circ }$

In $\Delta OPR,OR=OP$ (radii of the same circle)

$\therefore \angle OPR=\angle ORP=x$ (Say)

$\therefore \angle POR={180}^{\circ }-2x$

Similarly in $\Delta OQS$

OS = OQ

$\therefore \angle OSQ=\angle SQO=y$ (say)

$\therefore \angle SOQ={180}^{\circ }-2y$

POQ is a straight line,

$\angle POR+\angle ROS+\angle SOQ={180}^{\circ }$

$\Rightarrow {180}^{\circ }-2x+{42}^{\circ }+{180}^{\circ }-2y={180}^{\circ }$

$\Rightarrow {222}^{\circ }-2x-2y=0\Rightarrow 2(x+y)={222}^{\circ }$

$\therefore x+y={111}^{\circ }$ . . . (i)

In $\Delta PQT$,

$\Rightarrow \angle P+\angle Q+\angle T={180}^{\circ }$

$\Rightarrow \angle OPR+\angle SQO+\angle RTS={180}^{\circ }$

$x+y+\angle RTS={180}^{\circ }$

$\therefore \angle RTS={180}^{\circ }-(x+y)$

$={180}^{\circ }-{111}^{\circ }$ [from (i)]

$={69}^{\circ }$