In the figure below, lines AB and CD intersect at O. If
∠AOC+∠BOE=70∘ and
∠BOD=40∘,
find (2∠BOE).
60∘
∠AOC+∠BOE+∠COE=180∘ (linear pair of angle)
Given, ∠AOC+∠BOE=70∘
So, ∠COE=180∘–(∠AOC+∠COE)⇒180∘–70∘=110∘
Again, ∠COE+∠BOE+∠BOD=180∘ (linear pair of angle)
We know that, ∠COE=110∘ and ∠BOD=40∘
So, ∠BOE=180∘–∠COE−∠BOD⇒180∘–110∘−40∘=30∘
(2∠BOE)=60∘