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Question
Two lines AB and CD intersect at O. If `∠`AOC + `∠`COB + `∠`BOD = 270°, find the
measures of `∠`AOC, `∠`COB, `∠`BOD and `∠`DOA.
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Solution
Given: `∠`AOC + `∠`COB + `∠`BOP = 270°
To find: `∠`AOC, `∠`COB, `∠`BOD and `∠`DOA
Here, `∠`AOC + `∠`COB + `∠`BOD + `∠`AOD = 360° [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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