Question

PQRS is a cyclic quadrilateral and PQ is the diameter of the circle If$\angle QPR={35}^{\circ }$ then the value of $\angle PSR$ IS equal to

• A. ${125}^{\circ }$
• B. ${135}^{\circ }$
• C. ${145}^{\circ }$
• D. ${155}^{\circ }$

Answer 1

Answer: (A)

${125}^{\circ }$

$\Rightarrow$ $\angle QPR={35}^o$ [ Given ]

$\Rightarrow$ $\angle PRQ={90}^o$ [ Angle inscribed in semi-circle ]

$\Rightarrow$ $\angle QPR+\angle PRQ+\angle RQP={180}^o$ [ Sum of interior angles of triangle is ${180}^o$ ]

$\Rightarrow$ ${35}^o+{90}^o+\angle RQP={180}^o$

$\therefore$ ${125}^o+\angle RQP={180}^o$

$\therefore$ $\angle RQP={55}^o$

$\Rightarrow$ $\angle PSR+\angle RQP={180}^o$ [ Sum of pair of opposite angles of cyclic quadrilateral is ${180}^o$ ]

$\therefore$ $\angle PSR+{55}^o={180}^o$

$\therefore$ $\angle PSR={180}^o-{55}^o$

$\therefore$ $\angle PSR={125}^o$