If the figure below forms a linear pair,

Solution:

(i) ∠FOE = x, ∠DOE = y and ∠BOC = z

Since ∠AOF, ∠FOG is a linear pair

∠AOF + 30 = 180

∠AOF = 180 - 30

∠AOF = 150

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF = 150

30 + z + 30 + y + x = 150

x + y + z =150 - 30 - 30

x + y + z = 90  .... (1)

∠FOC = 90°

∠FOE + ∠EOD + ∠DOC = 90°

x + y + 30 = 90

x + y = 90 - 30

x + y =60 ... (2)

Substituting (2) in (1)

x + y + z = 90

60 + z = 90

z = 90 - 60 = 30

Given BOE = 90

∠BOC + ∠COD + ∠DOE = 90°

30 + 30 + DOE = 90

DOE = 90 - 60 = 30

DOE = x = 30

We also know that,

x + y = 60

y = 60 - x

y = 60 - 30

y = 30

Thus we have ∠FOE = 30, ∠COB = 30 and ∠DOE = 30

(ii) Right angles are ∠DOG, ∠COF, ∠BOF, ∠AOD

(iii) Adjacent complementary angles are (∠AOB, ∠BOD); ( ∠AOC, ∠COD); ( ∠BOC, ∠COE)

(iv) Adjacent supplementary angles are (∠AOB, ∠BOG); (∠AOC, ∠COG); (∠AOD, ∠DOG)

(v) Adjacent angles are (∠BOC, ∠COD); (∠COD, ∠DOE); (∠DOE, ∠EOF)