In the given figure, the lines AB, CD and EF intersect at a point O. If ∠BOD = x°, ∠AOE = 2x° and ∠COF = 90°, find ∠AOE and ∠AOC.

Given:

∠BOD = x°, ∠AOE = 2x°, ∠COF = 90°

In the figure, three straight lines (AB, CD, and EF) intersect at point O.

Therefore, vertically opposite angles are:

∠BOD = ∠AOC. Therefore, ∠AOC = x°

∠AOE = ∠BOF. Therefore, ∠BOF = 2x°

∠COF = ∠DOE. Therefore, ∠DOE = 90°

Since AB is a straight line, the angles ∠AOE, ∠EOD and ∠DOB lie on a straight line and their sum is 180°.

∴ ∠BOD + ∠DOE + ∠AOE = 180°

Substituting in the above equation, we get:

x°+ 90° + 2x = 180°

⇒ 3x° + 90° = 180°

⇒ 3x° = 180° - 90°

⇒ 3x° = 90°

⇒ x° = $\dfrac{90^{\circ}}{3}$

⇒ x° = 30°

Let's find ∠AOE and ∠AOC by substituting the value of x:

∠AOE = 2x° = (2 x 30)° = 60°

∠AOC = x° = 30°

∠AOE = 60° and ∠AOC = 30°.