Question

In triangle $ABC$, $BC$ is produced to $D$, $A$ & $D$ are joined. If $\angle BAC={60}^o$, $\angle ABC={45}^o$. Find $\angle ACD$

Answer 1

$\angle ACD=\angle BAC+\angle ABC$ [Exteriar angle property]

$\Rightarrow \angle ACD={60}^{\circ }+{45}^{\circ }$

$\Rightarrow {105}^{\circ }$