Question -
Answer -
In the figure,
∠AOF and ∠FOG form a linear pair
∠EOB = ∠FOC = 90°
∠DOC = ∠FOG = ∠AOB = 30°
(i) ∠BOE = 90°, ∠AOB = 30°
But ∠BOE + ∠AOB + ∠EOG = 180°
⇒ 30° + 90° + ∠EOG = 180°
∴∠EOG = 180° – 30° – 90° = 60°
But ∠FOG = 30°
∴ ∠FOE = 60° – 30° = 30°
∠COD = 30°, ∠COF = 90°
∴ ∠DOF = 90° – 30° = 60°
∠DOE = ∠DOF – ∠EOF
= 60° – 30° = 30°
∠BOC = BOE – ∠COE
= 90° – 30° – 30° = 90° – 60° = 30°
(ii) Right angles are,
∠AOD = 30° + 30° + 30° = 90°
∠BOE = 30° + 30° + 30° = 90°
∠COF = 30° + 30° + 30° = 90°
and ∠DOG = 30° + 30° + 30° = 90°
(iii) Pairs of adjacent complementaiy angles are ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠BOC, ∠COE
(iv) Pairs of adjacent supplementary angles are ∠AOB, ∠BOG; ∠AOC, ∠COG and ∠AOD, ∠DOG
(v) Pairs of adjacent angles are ∠BOC, ∠COD; ∠COD, ∠DOE and ∠DOE, ∠EOF.