Question

From the given triangle $ABC,$ find the measure of an $\angle ACB$.

Answer 1

Given, $\angle BAC={53}^o,\angle ACD={100}^o$

From the figure,
$\angle ACB+\angle ACD=180$ (Angles on a straight line)
$\angle ACB+100=180$
$\therefore \angle ACB={80}^{\circ }$