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Lines and Angles

Short Answer Type

1. Ray OE bisects ∠AOB and OF is the ray opposite to OE. Show that ∠FOB = ∠FOA.

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Rays OA, OB. OC, OD and OE have the common initial point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

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3. Prove “if two lines intersect each other, then the vertically opposite angles are equal.”

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4. In figure, OP bisects ∠AOC, OQ bisects ∠BOC and OP ⊥ OQ. Show that the points A, O and B are collinear.

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5.
In figure, if y = 20°, prove that the line AOB is a straight line.

∵ Sum of all the angle round a point is equal to 360°.
∴ y + (3x - 15) + (y + 5) + 2y + (4y + 10) + x = 360°
⇒ 4x + 8y = 360°
⇒ x + 2y = 90°
⇒ x + 2(20°) = 90°
⇒ x + 40° = 90°
⇒ x = 50°
Now, y + 3x - 15 + y + 5 = 3x + 2y - 10
= 3(50°) + 2(20°) - 10
= 150° + 40° - 10°
= 180°
∴ AOB is a straight line.

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