Question

Two lines AB and CD intersect at a point O such that $\angle BOC+\angle AOD={280}^{\circ }$, as shown in the figure. Find all the four angles.

Answer 1

We know that if two lines intersect, then the vertically-opposite angles are equal.

Let ∠BOC = ∠AOD = x°

Then,

x + x = 280

⇒ 2x = 280

⇒ x = 140°

∴∠BOC = ∠AOD = 140°

Also, let ∠AOC = ∠BOD = y°

We know that the sum of all angles around a point is 360°.

∴∠AOC + ∠BOC + ∠BOD + ∠AOD = 360°

⇒ y + 140 + y + 140 = 360°

⇒ 2y = 80°

⇒ y = 40°

Hence, ∠AOC = ∠BOD = 40°

∴∠BOC = ∠AOD = 140° and ∠AOC = ∠BOD = 40°