## LAOB LAO & LEOQ Fou form linear pair FOC – 98 and cool = cro4= = 30 LCOB o ML DOE Find m

Question

LAOB
LAO & LEOQ Fou form linear pair
FOC
– 98 and cool = cro4=
= 30
LCOB
o
ML DOE
Find mL FOE, ML COB

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2 weeks 2021-11-15T08:25:05+00:00 1 Answer 0 views 0

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1

maahira17

maahira17

01.05.2020

Math

Secondary School

+10 pts

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

nikitasingh79

nikitasingh79 Genius

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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:

∠AOB, ∠BOG

∠AOC, ∠COG

∠AOD, ∠DOG

(v) Three pairs of adjacent angles are:

∠BOC, ∠COD

∠COD, ∠DOE

∠DOE, ∠EOF