Question

In fig. given Line AB and CD intersect at O. If ∠AOC + ∠BOE =70˚ and ∠BOD =40˚, Find ∠COE .*
70˚
80˚
100˚
110˚​

Answer 1

Step-by-step explanation:

∠AOC + ∠BOE = 70°

∠AOC + ∠COE + ∠BOE = 180°

[ linear pair ]

So,

if ∠AOC + ∠BOE = 70°

so,

→ 70° + ∠COE = 180°

→ ∠COE = 180 - 70

→ ∠COE = 110°

.

∠BOD = ∠AOC [ Vertically Opposite Angles ]

.

Now,

→ ∠AOC + ∠COE + ∠BOE = 180°

→ 40° + 110° + ∠BOE = 180°

→ 150° + ∠BOE = 180°

→ ∠BOE = 180° - 150°

→ ∠BOE = 30°

.

∠BOD + ∠DOA = 180° [Liner Pair]

→ 40° + ∠DOA = 180°

→ ∠DOA = 180° - 40°

→ ∠DOA = 140°

Hence,

reflex angle ( ∠COE ) = ∠AOC + ∠DOE + ∠BOD + ∠BOE

reflex angle ( ∠COE ) = 40° + 140° + 40° + 30°

reflex angle ( ∠COE ) = 250°

Answer 2

Question :-

In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer :-

Since AB is a straight line,

∴ ∠AOC + ∠COE + ∠EOB = 180°

or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]

or ∠COE = 180° – 70° = 110°

∴ Reflex ∠COE = 360° – 110° = 250°

Also, AB and CD intersect at O.

∴∠COA = ∠BOD [Vertically opposite angles]

But ∠BOD = 40° [Given]

∴ ∠COA = 40°

Also, ∠AOC + ∠BOE = 70°

∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°

Thus, ∠BOE = 30° and reflex ∠COE = 250°.

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