Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measures of ∠AOC, ∠COB, ∠BOD, ∠DOA
Question:

Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measures of ∠AOC, ∠COB, ∠BOD, ∠DOA

 

Solution:

Given: ∠AOC + ∠COB + ∠BOD = 270°

To find: ∠AOC, ∠COB, ∠BOD, ∠DOA

Here, ∠AOC + ∠COB + ∠BOD = 270° [Complete angle]

⟹ 270 + AOD = 360

⟹ AOD = 360 – 270

⟹ AOD = 90

Now, AOD + BOD = 180         [Linear pair]

90 + BOD = 180

⟹ BOD = 180 – 90

⟹ BOD = 90

AOD = BOC = 90        [Vertically opposite angles]

BOD = AOC = 90        [Vertically opposite angles]

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