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Lines And Angles
Question
Rays OA, OB. OC, OD and OE have the common initial point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
Solution
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°.
Some More Questions From Lines and Angles Chapter
Ray OE bisects ∠AOB and OF is the ray opposite to OE. Show that ∠FOB = ∠FOA.
Rays OA, OB. OC, OD and OE have the common initial point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
Prove “if two lines intersect each other, then the vertically opposite angles are equal.”
In figure, OP bisects ∠AOC, OQ bisects ∠BOC and OP ⊥ OQ. Show that the points A, O and B are collinear.
In figure, if y = 20°, prove that the line AOB is a straight line.
In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex angle EOC.
In figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.
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.
If the complement of an angle is one-third of its supplement, find the angle.
In figure, find the value of x.
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Mock Test Series
Mock Test Series