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]