Question

In the given figure, AD divides ∠BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.

Answer 1

$$\begin{gathered} \angle BAC+\angle CAE=180°\left[\because BE\text{is a straight line}\right] \\ \Rightarrow \angle BAC+108°=180° \\ \Rightarrow \angle BAC=72° \end{gathered}$$

Now, divide $72°$ in the ratio 1 : 3.
$$\begin{gathered} \therefore a+3a=72° \\ \Rightarrow a=18° \\ \therefore a=18°\text{and}3a=54° \end{gathered}$$
Hence, the angles are 18^o and 54^o
$\therefore \angle BAD=18°\text{and}\angle DAC=54°$

Given,
$$\begin{gathered} AD=DB \\ \Rightarrow \angle DAB=\angle DBA=18° \end{gathered}$$

In $∆ABC$, we have:
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 72°+18°+x°=180° \\ \Rightarrow x°=90° \\ \therefore x=90 \end{gathered}$$