Question

In the given figure 'O' is the center of the circle. SP and TP are the two tangents at S and T respectively. $\angle SPT$ is ${50}^{\circ }$, the value of $\angle SQT$ is:

• A. 125°
• B. 65°
• C. 115°
• D. None of the above

Answer 1

The correct option is C

115°

$\angle OSP$ =$\angle OTP$ = ${90}^{\circ }${ radius is perpendicular to tangent in circle}
$\angle OSP+\angle SPT+\angle PTO+\angle TOS=360$
{ sum of all four angles of a triangle is ${360}^0$
${90}^0+{50}^0+{90}^0+\angle TOS={360}^0$
$\angle SOT$ = ${130}^{\circ }$
Now,
$\angle SOT+\angle SQT={360}^0$
{ sum of opposite angles angle sof a quadrilateral is ${360}^0$}
So,
${130}^0+\angle SQT={180}^0$
$\angle SQT={50}^0$