Question

In the given figure, PQ is a diameter and O is the center of the circle.

If $\angle PRT=\angle QTR={90}^{\circ }$

then which of the following statements are correct?

• A. $△PSR\sim △SQT$
• B. $\angle PSR=\angle OSQ$
• C. $\angle PSR=\angle SPO$
• D. $△OPS\sim △OQS$

Answer 1

Answer: (A), (B)

The correct options are
A

$△PSR\sim △SQT$

B

$\angle PSR=\angle OSQ$

$Let\angle PSR=x^{\circ }$

$\angle PSR=\angle PQS=x^{\circ }$....(Angles in alternate segment.)

$But\angle OSQ=\angle POQ=x^{\circ }$....(ΔOQS is isosceles.)

$\therefore \angle PSR=\angle OSQ=x^{\circ }$

$\angle OST={90}^{\circ }$....(Tangent ⊥ Radius at the [point of contact.)

$\angle QST={90}^{\circ }-x^{\circ }$

$\angle SQT=x^{\circ }$.....(Sum of angles of a triangle.)

$\angle RPS={90}^{\circ }-x^{\circ }$.....(Sum of angles of a triangle.)

$\therefore △PSR\sim △SQT$