Question

In the following figure, $\angle ABC={60}^{\circ },AC=CD$ and $\angle ADE={150}^{\circ }.$ Find the value of the $\angle BAC.$

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${70}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

Given:
$\angle ABC={60}^{\circ }AC=CD\angle ADE={150}^{\circ }$
$\angle ADE+\angle ADC={180}^{\circ }$(Linear pair)
${150}^{\circ }+\angle ADC={180}^{\circ }\angle ADC={180}^{\circ }-{150}^{\circ }={30}^{\circ }$
Since, $AC=CD\angle CAD=\angle CDA={30}^{\circ }$
(Angles opposite to equal sides are equal)

In$△ACD,$ by angle sum property,
$\angle ACD+\angle CDA+\angle CAD={180}^{\circ }\angle ACD+{30}^{\circ }+{30}^{\circ }={180}^{\circ }\angle ACD={180}^{\circ }-{30}^{\circ }-{30}^{\circ }={120}^{\circ }\angle ACD={120}^{\circ }$

$\angle ACD$ acts as the exterior angle for the $△ABC.$
So, by exterior angle property,
$\angle ACD=\angle ABC+\angle BAC{120}^{\circ }={60}^{\circ }+\angle BAC\angle BAC={60}^{\circ }$