Question

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If $\angle QPR={67}^{o}and\angle SPR={72}^o$, then $\angle QRS=$

• A. ${41}^o$
• B. ${23}^o$
• C. ${67}^o$
• D. ${18}^o$

Answer 1

The correct option is A

${41}^o$

PQRS is a cyclic quadrilateral with centre O and $\angle QPR={67}^o$, $\angle SPR={72}^o$.

$\therefore \angle QPS={67}^o+{72}^o={139}^o$

$\because \angle QPS+\angle QRS={180}^o$ (Sum of opposite angles of a cyclic quadrilateral)

$\Rightarrow {139}^o+\angle QRS={180}^o$

$\Rightarrow \angle QRS={180}^o-{139}^o={41}^o$