Question

In the figure, $CDE$ is a straight line and $A,B,C$ and $D$ are points on the circle. $\angle BCD={44}^{\circ },$ find the value of $x.$

• A. ${44}^{\circ }$
• B. ${68}^{\circ }$
• C. ${90}^{\circ }$
• D. ${56}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

$\angle CDB=\frac{1}{2}({180}^{\circ }-{44}^{\circ })=\frac{1}{2}\times {136}^{\circ }={68}^{\circ }$ ($\therefore$ $BCD$ is an isos. $△$)
$\angle BAD={180}^{\circ }-{44}^{\circ }={136}^{\circ }$ (opp. $\angle$s of a cyclic quad. are supp.)
$\therefore$ $\angle ADB=\frac{1}{2}({180}^{\circ }-{136}^{\circ })=\frac{1}{2}\times {44}^{\circ }={22}^{\circ }$ ($\because$ $BAD$ is an isos. $△$)
$\therefore$ $\angle ADC=\angle ADB+\angle BDC={22}^{\circ }+{68}^{\circ }={90}^{\circ }$
$\Rightarrow$ $x=\angle ADE={180}^{\circ }-\angle ADC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$ ($\because$ $EDC$ is a st. line)