Question

In the figure, the sides BC,CA and AB of a $\Delta ABC$, have been produced to D, E and F respectively. If $\angle ACD={105}^{\circ }$ and $\angle EAF={45}^{\circ }$, find all the angles of the $\Delta ABC$.

Answer 1

In $\Delta ABC$. sides BC, CA and are produced to D,E and F respectively.

Also, $\angle ACD$=${105}^{\circ }$ and $\angle EAF={45}^{\circ }$

$\angle ACD+\angle ACB={180}^{\circ }$ (Linear pair)

$\Rightarrow {105}^{\circ }+\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{105}^{\circ }={75}^{\circ }$

$\angle BAC=\angle EAF={45}^{\circ }$ (Vertically opposite angles)

But $\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$

$\Rightarrow {45}^{\circ }+\angle ABC+{75}^{\circ }={180}^{\circ }$

$\Rightarrow {120}^{\circ }+\angle ABC={180}^{\circ }$

$\Rightarrow \angle ABC={180}^{\circ }-{120}^{\circ }$

$\therefore \angle ABC={60}^{\circ }$

Hence $\angle ABC={60}^{\circ },\angle ACB={75}^{\circ }$ and $\angle BAC={45}^{\circ }$