Question

4) Two lines and AB and CD intersect at O. If AOC + BOD equal 250°, find AOC, AOD,
BOC.
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Answer 1

Answer:

Lets draw the figure first. (To see figure, see file attached below)

∠AOC+∠BOD=250°.

But, ∠BOD=∠AOC (VERTICALLY OPPOSITE ANGLES ARE EQUAL)

Thus, ∠AOC+∠AOC=250°........(∵ ∠BOD=∠AOC)

2∠AOC=250°

∠AOC=250÷2

∠AOC=125°.

Now, we can use linear pair.

∠AOD+∠AOC=180°.........(Linear pair)

∠AOD=125°=180°.......(∵∠AOC=125°)

∠AOD=55°.

And,

∠AOC+∠BOC=180°..............(Linear pair)

125°+∠BOC=180..........(∠AOC=125°)

∠BOC=180-125

∠BOC=55°.

THUS, ∠AOC=125°, ∠AOD=55°, ∠BOC=55°.

HOPE THIS HELPS :D

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°

o r (∠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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