Question

In the given figure, three lines AB, CD and EF intersect at a point O such that $\angle AOE={35}^{\circ }$ and $\angle BOD={40}^{\circ }$. Find the measure of $\angle AOC,\angle BOF,\angle COF$ and $\angle DOE$.

Answer 1

In the given figure,

∠AOC = ∠BOD = 40${}^o$ (Vertically opposite angles)

∠BOF = ∠AOE = 35${}^o$ (Vertically opposite angles)

Now, ∠EOC and ∠COF form a linear pair.

∴ ∠EOC + ∠COF = 180${}^o$

⇒ (∠AOE + ∠AOC) + ∠COF = 180${}^o$

⇒ 35${}^o$ + 40${}^o$ + ∠COF = 180${}^o$

⇒ 75${}^o$ + ∠COF = 180${}^o$

⇒ ∠COF = 180${}^o$ − 75${}^o$ = 105${}^o$

Also, ∠DOE = ∠COF = 105${}^o$

(Vertically opposite angles)