Question

In the given, PQ is a diameter. Chord SR is parallel to PQ. Given that $\angle PQR={58}^{\circ }$ Calculate: $\angle STP$

• A. ${104}^{\circ }$
• B. ${208}^{\circ }$
• C. ${18}^{\circ }$
• D. ${148}^{\circ }$

Answer 1

The correct option is D

${148}^{\circ }$

Join PR,

In $\Delta PQR$

$\angle PRQ={90}^{\circ }$ (Angle in a semi - circle)

$\therefore \angle RPQ+\angle RQP={90}^{\circ }$

$\Rightarrow \angle RPQ+{58}^{\circ }={90}^{\circ }$

$\Rightarrow \angle RPQ={90}^{\circ }-{58}^{\circ }={32}^{\circ }$ . . . (i)

$\therefore SR||PQ$ (given)

$\therefore \angle SRP=\angle RPQ$ (alternate angles)

$={32}^{\circ }$ (from (i))

In cyclic quadrilateral. PRST,

$\angle STP+\angle SRP={180}^{\circ }$

$\Rightarrow \angle STP+{32}^{\circ }={180}^{\circ }$

$\therefore \angle STP={180}^{\circ }-{32}^{\circ }={148}^{\circ }$