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]

Concept: Concept of Parallel Lines

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