Question

Given the triangle $ABC$ with angles shown in the diagram above, then $m\angle ACD=$

• A. ${100}^{\circ }$
• B. ${130}^{\circ }$
• C. ${120}^{\circ }$

Answer 1

Answer: (A), (C)

${120}^{\circ }$


In the diagram shown for $△ABC$:
$m\angle B={60}^{\circ };m\angle A=m\angle C=x^{\circ }$
Now, according to the angle sum property:
$m\angle A+m\angle B+m\angle C={180}^{\circ }$
$\Rightarrow x^{\circ }+{60}^{\circ }+x^{\circ }={180}^{\circ }$
$\Rightarrow 2x^{\circ }={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\Rightarrow 2x^{\circ }={120}^{\circ }\Rightarrow x^{\circ }={60}^{\circ }$
$\therefore \angle A=x^{\circ }={60}^{\circ }$
Now, $\angle ACD$ is an external angle.
$\Rightarrow m\angle ACD=m\angle A+m\angle B={120}^{\circ }$