Question

In the given figure, the side $BC$ of $△ABC$ has been produced on the left-hand side from $B$ to $D$ and on the right -hand side from $C$ to $E$. if $\angle ABD={106}^{\circ }$ and $\angle ACE={118}^{\circ }$, find the measure of each angle of the triangle.

Answer 1

From the figure we know that $\angle DBA$ and $\angle ABC$ form a linear pair of angles
So we get
$\angle DBA+\angle ABC={180}^{\circ }$
By substituting the values
${106}^{\circ }+\angle ABC={180}^{\circ }$
On further calculation
$\angle ABC={180}^{\circ }-{106}^{\circ }$
By subtraction
$\angle ABC={74}^{\circ }$
From the figure we know that $\angle ACB$ and $\angle ACE$ form a linear pair of angles
So we get
$\angle ACB+\angle ACE={180}^{\circ }$
By substituting the values
$\angle ACB+{118}^{\circ }={180}^{\circ }$
On further calculation
$\angle ACB={180}^{\circ }-{118}^{\circ }$
By subtraction
$\angle ACB={62}^{\circ }$
We know that the sum of all the angles in a triangle is ${180}^{\circ }$.
So we can write it as
$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$
By substituting the values
${74}^{\circ }+{62}^{\circ }+\angle BAC={180}^{\circ }$
On further calculation
$\angle BAC={180}^{\circ }-{74}^{\circ }-{62}^{\circ }$
By subtraction
$\angle BAC={180}^{\circ }-{136}^{\circ }$
$\angle BAC={44}^{\circ }$
Therefore, the measure of each angle of the triangle is $\angle A={44}^{\circ },\angle B={74}^{\circ }$ and $\angle C={62}^{\circ }$.