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

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°

 

Solution:

(c) 80°

We have :

$\angle A O C+\angle C O D+\angle B O D=180^{\circ} \quad$ [Since $A O B$ is a straight line ]

$\Rightarrow 3 x-10+50+x+20=180$

$\Rightarrow 4 x=120$

 

$\Rightarrow x=30$

$\therefore \angle A O C=[3 \times 30-10]^{\circ}$

 

$\Rightarrow \angle A O C=80^{\circ}$

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