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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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2.
Rays OA, OB. OC, OD and OE have the common initial point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

Construction. Draw a ray OP opposite to ray OA.
Proof. ∠AOB + ∠BOC + ∠COP = 180° ...(1)
| ∵ A straight angle = 180°
∠POD + ∠DOE + ∠EOA = 180° ...(2)
| ∵ A straight angle = 180°
Adding (1) and (2), we get
∠AOB + ∠BOC + (∠COP + ∠POD) + ∠DOE + ∠EOA = 180° + 180° = 360°
⇒ ∠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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In figure, if y = 20°, prove that the line AOB is a straight line.

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6. In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex angle EOC.

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7. In figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.

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8. In figure, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ, respectively. If ∠POS = x, find ∠ROT.

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9. If the complement of an angle is one-third of its supplement, find the angle.

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10. In figure, find the value of x.

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Previous Year Papers

Test Series

Test Yourself

Short Answer Type

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

2.
Rays OA, OB. OC, OD and OE have the common initial point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.