Question

Question 41

In $\Delta ABC$, $\angle A={50}^{\circ },\angle B={70}^{\circ }$ and bisector of $\angle C$ meets AB at D as shown in the given figure. Measure of $\angle ADC$ is

a) ${50}^{\circ }$
b) ${100}^{\circ }$
c) ${30}^{\circ }$
d) ${70}^{\circ }$

Answer 1

In $\Delta ADC$,
$\angle ADC+\angle DAC+\angle ACD={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow \angle ADC+{50}^{\circ }+\angle ACD={180}^{\circ }$ $[\because \angle DAC={50}^{\circ }]$
$\Rightarrow \angle ACD={130}^{\circ }-\angle ADC\dots \dots \left(i\right)$
In $\Delta DBC,\angle ADC=\angle DBC+\angle BCD$
[$\because$ exterior angle is equal to sum of opposite interior angles]
$\Rightarrow \angle ADC={70}^{\circ }+\angle ACD$ $[\because \angle ACD=\angle BCD]$
$\Rightarrow \angle ADC={70}^{\circ }+{130}^{\circ }-\angle ADC$ [from Eq. (i)]
$\Rightarrow \angle ADC={200}^{\circ }-\angle ADC\Rightarrow 2\angle ADC={200}^{\circ }\Rightarrow \angle ADC=\frac{{200}^{\circ }}{2}\Rightarrow \angle ADC={100}^{\circ }$