Question

In the given figure, AD divides $\angle BAC$ in the ratio 1:3. Find the value of X and $\angle ADC$ (in degrees).

Answer 1

$\angle EAC=\angle ABC+X$ (Exterior angle property)
${120}^{\circ }={100}^{\circ }+X$
So, $X={20}^{\circ }$

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

$\angle BAC+X+{100}^{\circ }={180}^{\circ }$
Hence,
$\angle BAC={60}^{\circ }$
Therefore,
$\angle DAC=\frac{3}{4}\angle BAC={45}^{\circ }$
(because $\angle BAD$ and $\angle DAC$ are in ratio 1:3)

$\angle ADC={180}^{\circ }-({45}^{\circ }+20)={115}^{\circ }$