Question : In $\triangle {ABC}$, D is a point on BC such that $\angle {ADB}=2 \angle {DAC}, \angle {BAC}=70^{\circ}$ and $\angle {B}=56^{\circ}$. What is the measure of $\angle A D C$?
Option 1: $72^{\circ}$
Option 2: $54^{\circ}$
Option 3: $74^{\circ}$
Option 4: $81^{\circ}$
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Correct Answer: $72^{\circ}$
Solution :
Let $\angle DAC = x$
$\angle ADB = 2x$
$\angle BAC = 70^\circ$
$\angle BAD = 70^\circ - x$
In $\triangle ABD $
$⇒56^\circ + 2x^\circ + 70^\circ - x = 180^\circ $
$⇒ x = 54^\circ$
$⇒2x = 108^\circ$
Now, By linear pair of angles
$\angle ADB + \angle ADC = 180^\circ $
$⇒ 2x + \angle ADC = 180^\circ$
$⇒ 108^\circ + \angle ADC = 180^\circ $
$⇒ \angle ADC = 180^\circ - 108^\circ$
$⇒ \angle ADC = 72^\circ$
Hence, the correct answer is $72^\circ$.
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