In the given figure ; AOC-is a straight line. If angle AOB = 50°, ang

1.	In the given figure ; AOC-is a straight line. If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of :(i) angle BOC(ii) angle EOD(iii) obtuse angle BOD(iv) reflex angle BOD(v) reflex angle COE.
Answer» (i) ∠AOB + ∠BOC = 180° (Linear pairs of angle) ⇒ 50° +∠BOC = 180° ⇒ ∠BOC = 180° – 50° = 130° ⇒ ∠BOC = 130° (ii) ∠EOD + ∠COD = 90° (∵AOE = 90°) ⇒ ∠EOD + 25° = 90° ⇒ ∠EOD + 25° = 90° ⇒ ∠EOD = 90° – 25° ⇒ ∠EOD = 65° (iii) ∠BOD = ∠BOC + COD = 130° + 25° = 155° (iv) Reflex ∠BOD = 360° – ∠BOD = 360°- 155° = 205° (v) Reflex ∠COE = 360° – ∠COE = 360° (∠COD + ∠EOD) = 360° – (25° + 65°) = 360° – 90° = 270°

In the given figure ; AOC-is a straight line. If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of :(i) angle BOC(ii) angle EOD(iii) obtuse angle BOD(iv) reflex angle BOD(v) reflex angle COE.

Answer»

(i) ∠AOB + ∠BOC = 180° (Linear pairs of angle)

⇒ 50° +∠BOC = 180°

⇒ ∠BOC = 180° – 50° = 130°

⇒ ∠BOC = 130°

(ii) ∠EOD + ∠COD = 90° (∵AOE = 90°)

⇒ ∠EOD + 25° = 90°

⇒ ∠EOD = 90° – 25°

⇒ ∠EOD = 65°

(iii) ∠BOD = ∠BOC + COD

= 130° + 25° = 155°

(iv) Reflex ∠BOD = 360° – ∠BOD

= 360°- 155° = 205°

(v) Reflex ∠COE = 360° – ∠COE

= 360° (∠COD + ∠EOD)

= 360° – (25° + 65°)

= 360° – 90° = 270°