Question

In $ ∆\mathrm{ABC}$, $ \angle \mathrm{A}=50°$, $ \angle \mathrm{B}=70°$ and the bisector of $ \angle \mathrm{C}$ meets $ \mathrm{AB}$ in $ \mathrm{D}$ (figure). Measure of $ \angle \mathrm{ADC}$ is

• A. $50°$
• B. $100°$
• C. $30°$
• D. $70°$

Answer 1

The correct option is B

$100°$

Step 1: Find $\angle \mathrm{ACB}$:

Given: $\angle \mathrm{A}=50°,\angle \mathrm{B}=70°$

$$\begin{gathered} In∆ABC \\ \angle A+\angle B+\angle C=180°[\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}\mathrm{triangle}\mathrm{is}180°] \\ \Rightarrow 50+70+\angle C=180 \\ \Rightarrow 120+\angle C=180 \\ \Rightarrow \angle C=180-120 \\ \therefore \angle C=60° \end{gathered}$$

Step 2: Find the required angle:

$$\begin{gathered} In∆ADC \\ \angle A+\angle ADC+\angle ACD=180°[\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}\mathrm{triangle}\mathrm{is}180°] \\ \Rightarrow 50+\angle \mathrm{ADC}+30=180[\mathrm{DC}\mathrm{is}\mathrm{the}\mathrm{bisector}\mathrm{of}\angle \mathrm{C}] \\ \Rightarrow 80+\angle \mathrm{ADC}=180 \\ \Rightarrow \angle \mathrm{ADC}=180-80 \\ \therefore \angle \mathrm{ADC}=100° \end{gathered}$$

Hence option (B) is the correct answer.