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.

#### Solution

(i) `∠`*FOE *= *x*, `∠`*DOE *= *y *and `∠`*B**O**C *= *z *sat

Since `∠`*AOF *, `∠`*FOG *is Linear pair

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

⇒ `∠`*AOF *= 180° - 30°

*⇒ `∠`AOF *= 150°

⇒ `∠` *A**O**B *+ `∠`*B**O**C *+ `∠`*C**O**D *+ `∠`*DO**E *+ `∠`*E**O**F *= 150°

⇒ 30° + *z *+ 30° + *y *+ *x *= 150°

*⇒ x *+ *y *+ *z *= 150° - 30° - 30°

*⇒ x *+ *y *+ *z *= 90° .....(1)

Now `∠`*F**O**C *= 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 `∠`*B**O**E *= 90°

*⇒*`∠`*BOC *+ `∠`*COD *+ `∠`*DOE *= 90°

*⇒ *30° + 30° + `∠`*DOE *= 90°

*⇒ *`∠`*DOE *= 90° - 60° = 30°

∴ `∠`*DO**E *= *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*, `∠`*C**OD*;

*`∠`C**OD*, `∠`*DOE*;

*`∠`DOE*, `∠`*EOF *,