Question

In the given figure $\angle BAC$ measures ${30}^{\circ }$, $\angle ABC$ measures $110^{\circ }$, and points B, C, and D are collinear. Find the measure of $\angle ACD$.

• A. ${150}^{\circ }$
• B. ${140}^{\circ }$
• C. ${130}^{\circ }$
• D. ${120}^{\circ }$
• E. ${110}^{\circ }$

Answer 1

The correct option is B ${140}^{\circ }$

In $△ABC$

$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$ [By sum of angle property]

$\Rightarrow {30}^{\circ }+{110}^{\circ }+\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{110}^{\circ }$

$\Rightarrow \angle ACB={40}^{\circ }$

Points $B,C,D$ are collinear

$\therefore \angle ACB+\angle ACD={180}^{\circ }$

$\Rightarrow {40}^{\circ }+\angle ACD={180}^{\circ }$

$\Rightarrow \angle ACD={180}^{\circ }-{40}^{\circ }$

$\Rightarrow \angle ACD={140}^{\circ }$