If Below Fig, ∠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. - Mathematics

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If below fig, ∠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.

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Solution

(i)  `∠`FOE = x, `∠`DOE = y and `∠`BOC = z sat

Since `∠`AOF , `∠`FOG is Linear pair

⇒`∠`AOF + 30° = 180°              [`∠`AOF + `∠`FOG = 180° and `∠`FOG = 30°]

⇒ `∠`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)

Now `∠`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° Þ z = 90° - 60° = 30°

i.e., `∠`BOC = 30°

Given `∠`BOE = 90°

`∠`BOC + `∠`COD + `∠`DOE = 90°

⇒ 30° + 30° + `∠`DOE = 90°

⇒ `∠`DOE = 90° - 60° = 30°

∴ `∠`DOE = x = 30°

Now, also we have

x + y = 60°

⇒ y = 60° - x = 60° - 30° = 30°

`∠`FOE = 30

(ii) Right angles are

`∠`DOG, `∠`COF , `∠`BOF , `∠`AOD

(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

`∠`BOC, `∠`COD;

`∠`COD, `∠`DOE;

`∠`DOE, `∠`EOF ,

Concept: Pairs of Angles
  Is there an error in this question or solution?
Chapter 10: Lines and Angles - Exercise 10.2 [Page 16]

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RD Sharma Mathematics for Class 9
Chapter 10 Lines and Angles
Exercise 10.2 | Q 18 | Page 16

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