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In the given below fig, rays OA, OB, OC, OP and 0E have the common end point O. Show

that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

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#### Solution 1

Given that

Rays *OA*, *OB*, *OD *and *O**E *have the common end point O.

A ray of opposite to OA is drawn

Since `∠`*AOB*, `∠`*BOF *are linear pairs

*`∠`AOB *+ `∠`*BOF *= 180°

*`∠`AOB *+ `∠`*BOC *+ `∠`*COF *= 180°

Also

*`∠`AOE*, `∠`*EOF *are linear pairs

*`∠`AOE *+ `∠`*EOF *= 180°

*`∠`AOE *+ `∠`*DOF *+ `∠`*DOE *= 180°

By adding (1) and (2) quations we get

*`∠`A**O**B *+ `∠`*B**O**C *+ `∠`*C**O**F *+ `∠`*A**O**E *+ `∠`*DO**F *+ `∠`*DO**E *= 360°

*`∠`A**O**B *+ `∠`*B**O**C *+ `∠`*C**O**D *+ `∠`*DO**E *+ `∠`*E**O**A *= 360°

Hence proved.

#### Solution 2

Let us draw AOXa straight line.

∠AOE,∠DOE and ∠DOXform a linear pair. Thus, their sum should be equal to180°.

Or, we can say that:

∠AOE +∠DOE +∠DOX = 180° (I)

Similarly,, ∠AOB,∠BOC and ∠COXform a linear pair. Thus, their sum should be equal to180°.

Or, we can say that:

∠AOB +∠BOC+ ∠COX = 180° (II)

On adding (I) and (II), we get:

∠AOB +∠BOC + ∠COX +∠DOX +∠AOE +∠DOE = 180°+180°

∠AOB +∠BOC + ∠COD +∠AOE +∠DOE = 360°

Hence proved.

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