Question

In Figure,, the sides BC, CA and BA 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 a $\Delta ABC,\angle BACand\angle EAF$ are vertically opposite angles.
Hence, we can write as
$\angle BAC=\angle EAF={45}^{\circ }$
Considering the exterior angle property, we have
$\angle BAC+\angle ABC=\angle ACD={105}^{\circ }$
On rearranging we get
$\angle ABC={105}^{\circ }-{45}^{\circ }={60}^{\circ }$
We know that the sum of angles in a triangle is 180o
$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$
$\angle ACB={75}^{\circ }$
Therefore, the angles are ${45}^{\circ },{60}^{\circ }and{75}^{\circ }$