Question

Two lines AB and CD intersect at a point O such that ∠BOC + ∠AOD = 280°, as shown in the figure. Find all the four angles.

Answer 1

From the figure we know that ∠AOD and ∠BOC are vertically opposite angles

∠AOD = ∠BOC

It is given that

∠BOC + ∠AOD = 280^o

We know that ∠AOD = ∠BOC

So it can be written as

∠AOD + ∠AOD = 280^o

On further calculation

2 ∠AOD = 280^o

By division

∠AOD = 280/2

∠AOD = ∠BOC = 140^o

From the figure we know that ∠AOC and ∠AOD form a linear pair

So it can be written as

∠AOC + ∠AOD = 180^o

Substituting the values

∠AOC + 140^o = 180^o

On further calculation

∠AOC = 180^o – 140^o

By subtraction

∠AOC = 40^o

From the figure we know that ∠AOC and ∠BOD are vertically opposite angles

∠AOC = ∠BOD = 40^o

Therefore, ∠AOC = 40^o, ∠BOC = 140^o, ∠AOD = 140^o and ∠BOD = 40^o