(iii) Name three pairs of adjacent complementary angles.

(iv) Name three pairs of adjacent supplementary angles.

(v) Name three pairs of adjacent angles.

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Given : ∠AOF and ∠FOG form a linear pair.

∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°

Proof:

(i) Let ∠FOE = x , ∠DOE = y and ∠COB = z

∠AOF + ∠FOG = 180° (Linear pair)

∠AOF + 30° = 180°

∠AOF = 180° – 30°

∠AOF = 150°

But

∠AOF = ∠AOB + ∠BOC + ∠DOC + ∠DOE + ∠EOF

150° = 30° + z + 30° + y + x

150° = 60° + z + + y + x

150° – 60° = x + y + z

x + y + z = 90° …………..(1)

Now,

∠FOC = ∠FOE + ∠DOE + ∠DOC

90° = ∠FOE + ∠EOD + ∠DOC

90° = x + y + 30°

90° – 30° = x + y

x + y = 60° ……………. (2)

Put this value of x + y in eq 1, we obtain

x + y + z = 90°

60° + z = 90°

z = 90° – 60°

z = 30°

Then, ∠COB = 30°

Now,

∠EOB = ∠BOC + ∠COD + ∠DOE

90° = 30° + 30° + ∠DOE

90° = 60° + ∠DOE

∠DOE = 90° – 60°

∠DOE = 30°

But,

∠FOC = ∠FOE + ∠DOE + ∠DOC

90° = ∠FOE + ∠DOE + ∠DOC

90° = x + 30° + 30°

[∠DOE = 30°]

90° = x + 60°

x = 90° – 60°

x = 30°

∠FOE = x = 30°

Hence, the measures of ∠FOE, ∠COB and ∠DOE is 30°.

(ii) Right angles are:

∠AOD , ∠DOG, ∠FOC, ∠BOE

(iii) Three pairs of adjacent complementary angles are:

∠AOB, ∠BOD

∠AOC, ∠COD

∠BOC, ∠COE

(iv) Three pairs of adjacent supplementary angles are:

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maahira17

maahira17

01.05.2020

Math

Secondary School

+10 pts

Answered

In Fig. 8.44, ∠AOF and ∠FOG form a linear pair.

∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°

(i) Find the measures of ∠FOE, ∠COB and ∠DOE.

(ii) Name all the right angles.

(iii) Name three pairs of adjacent complementary angles.

(iv) Name three pairs of adjacent supplementary angles.

(v) Name three pairs of adjacent angles.

2

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Answers

THE BRAINLIEST ANSWER!

nikitasingh79

nikitasingh79 Genius

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Verified Answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but Verified Answers are the finest of the finest.

Given : ∠AOF and ∠FOG form a linear pair.

∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°

Proof:

(i) Let ∠FOE = x , ∠DOE = y and ∠COB = z

∠AOF + ∠FOG = 180° (Linear pair)

∠AOF + 30° = 180°

∠AOF = 180° – 30°

∠AOF = 150°

But

∠AOF = ∠AOB + ∠BOC + ∠DOC + ∠DOE + ∠EOF

150° = 30° + z + 30° + y + x

150° = 60° + z + + y + x

150° – 60° = x + y + z

x + y + z = 90° …………..(1)

Now,

∠FOC = ∠FOE + ∠DOE + ∠DOC

90° = ∠FOE + ∠EOD + ∠DOC

90° = x + y + 30°

90° – 30° = x + y

x + y = 60° ……………. (2)

Put this value of x + y in eq 1, we obtain

x + y + z = 90°

60° + z = 90°

z = 90° – 60°

z = 30°

Then, ∠COB = 30°

Now,

∠EOB = ∠BOC + ∠COD + ∠DOE

90° = 30° + 30° + ∠DOE

90° = 60° + ∠DOE

∠DOE = 90° – 60°

∠DOE = 30°

But,

∠FOC = ∠FOE + ∠DOE + ∠DOC

90° = ∠FOE + ∠DOE + ∠DOC

90° = x + 30° + 30°

[∠DOE = 30°]

90° = x + 60°

x = 90° – 60°

x = 30°

∠FOE = x = 30°

Hence, the measures of ∠FOE, ∠COB and ∠DOE is 30°.

(ii) Right angles are:

∠AOD , ∠DOG, ∠FOC, ∠BOE

(iii) Three pairs of adjacent complementary angles are:

∠AOB, ∠BOD

∠AOC, ∠COD

∠BOC, ∠COE

(iv) Three pairs of adjacent supplementary angles are:

∠AOB, ∠BOG

∠AOC, ∠COG

∠AOD, ∠DOG

(v) Three pairs of adjacent angles are:

∠BOC, ∠COD

∠COD, ∠DOE

∠DOE, ∠EOF

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