Question

In the given figure, the side BC of ∆ ABC has been produced on both sides−on the left to D and on the right to E. If ∠ABD = 106° and ∠ACE = 118°, find the measure of each angle of the triangle.

Answer 1

Side BC of triangle ABC is produced to D.

$$\begin{gathered} \therefore \angle ABC=\angle A+\angle C \\ \Rightarrow 106°=\angle A+\angle C...\left(i\right) \end{gathered}$$

Also, side BC of triangle ABC is produced to E.

$$\begin{gathered} \angle ACE=\angle A+\angle B \\ \Rightarrow 118°=\angle A+\angle B...\left(ii\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding}\left(i\right)\mathrm{and}\left(ii\right),\mathrm{we}\mathrm{get}: \\ \angle A+\angle A+\angle B+\angle C=\left(106+118\right)° \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(\angle A+\angle B+\angle C\right)+\angle A=224°\left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow 180°+\angle A=224° \\ \Rightarrow \angle A=44° \end{gathered}$$

$$\begin{gathered} \therefore \angle B=118°-\angle A\left[\text{Using}\left(ii\right)\right] \\ \Rightarrow \angle B=\left(118-44\right)° \\ \Rightarrow \angle B=74° \end{gathered}$$

And,

$$\begin{gathered} \angle C=106°-\angle A\left[\text{Using}\left(i\right)\right] \\ \Rightarrow \angle C=\left(106-44\right)° \\ \Rightarrow \angle C=62° \end{gathered}$$