In the figure given below, find the measure of $\angle \mathrm{ADC}$. 

In the figure given alongside, the measure of $\angle \mathrm{ADC}$ is 50°.

Given a figure.

As sum of angle on a straight line is supplementary,

$\angle \mathrm{ACB}+\angle \mathrm{ACD}={180}^{\circ }$

$\begin{array}{l}{100}^{\mathrm{o}}+\angle \mathrm{ACD}={180}^{\mathrm{o}}\\ \angle \mathrm{ACD}={180}^{\mathrm{o}}-{100}^{\mathrm{o}}\\ \angle \mathrm{ACD}={80}^{\mathrm{o}}\end{array}$  

Now, we know that the sum of the all-interior angles of a triangle is ${180}^{\mathrm{o}}$. 

$\begin{array}{l}\angle \mathrm{ADC}+{50}^{\mathrm{o}}+{80}^{\mathrm{o}}={180}^{\mathrm{o}}\\ \angle \mathrm{ADC}={180}^{\mathrm{o}}-{130}^{\mathrm{o}}\\ \angle \mathrm{ADC}={50}^{\mathrm{o}}\end{array}$ 

Therefore, the measure of $\angle \mathrm{ADC}$  is ${50}^{\mathrm{o}}$.