Question

In the given figure, find the value of $x$.

• A. ${25}^{\circ }$
• B. ${30}^{\circ }$
• C. ${35}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

$\angle TQS=180-\angle PQT-\angle SQR=180–70–50=60$

WKT $OQ\perp PR$

$\angle OQS=90-\angle SQR=90–50=40$

In $△OQS$

$\angle OQA=\angle OSQ=40$

$\angle QOS=180-\angle OQS-\angle OSQ=180–40–40=100$

Now,

$\angle QTS=\frac{1}{2}\angle QOS$ because angle made but n arc at the center is twice the angle made by the arc on any point of circle

$\angle QTS=\frac{100}{2}=50$

In $△QTS$

$\angle TSQ=180–50-60=70$

$\angle TSO+\angle OSQ=70$

$x=70–40={30}^{\circ }$