Question

In the figure given below, AB is a diameter of the semi-circle APQB with centre O. $\angle POQ={48}^{\circ }$. AQ cuts BP at X, calculate $\angle AXP$.

• A. ${50}^{\circ }$
• B. ${55}^{\circ }$
• C. ${66}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is C ${66}^{\circ }$

$\angle PBQ=\frac{1}{2}\left({48}^{\circ }\right)={24}^{\circ }$
($\angle$ at centre = $2\times \angle$ at circumference on same PQ)
$\angle AQB={90}^{\circ }$ ($\angle$ in semi-circle)
$\angle AXP=\angle BXQ={180}^{\circ }-{90}^{\circ }-{24}^{\circ }$ $={66}^{\circ }$ ($\angle$ sum of $\Delta$ )