Question

In the given figure, AOB is a straight line. If ∠AOC = (3x − 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer 1

(c) 80°

We have :
$$\begin{gathered} \angle AOC+\angle COD+\angle BOD=180°\left[\text{Since}\mathit{\text{AOB}}\text{is a straight line}\right] \\ \Rightarrow 3x-10+50+x+20=180 \\ \Rightarrow 4x=120 \\ \Rightarrow x=30 \end{gathered}$$

$$\begin{gathered} \therefore \angle AOC={\left[3\times 30-10\right]}^{°} \\ \Rightarrow \angle AOC=80° \end{gathered}$$