Question - 15 : - In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT. [NCERT]

Answer - 15 : -

Ray OR stands on a line POQ forming ∠POS and ∠QOS

OR and OT the angle bisects of ∠POS and ∠QOS respectively. ∠POS = x

But ∠POS + ∠QOS = 180° (Linear pair)

⇒ x + ∠QOS = 180°

⇒ ∠QOS = 180° – x

∵ OR and OT are the bisectors of angle

Question - 16 : - In the figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7. Find all the angles. [NCERT]

Answer - 16 : -

Lines PQ and PR, intersect each other at O

∵ Vertically opposite angles are equal

∴ ∠POR = ∠QOS and ∠ROQ = ∠POS

∠POR : ∠ROQ = 5:7

Let ∠POR = 5x and ∠ROQ = 7x

But ∠POR + ∠ROQ = 180° (Linear pair)

∴ 5x + 7x = 180° ⇒ 12x = 180°

⇒ x =

= 15°

∴ ∠POR = 5x = 5 x 15° = 75°

and ∠ROQ = 7x = 7 x 15° = 105°

But ∠QOS = POR = 75° (Vertically opposite angles)

and ∠POS = ∠ROQ = 105°

Question - 17 : - In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.

Answer - 17 : -

In the figure,

∠AOC + ∠BOC = 180° (Linear pair)

⇒ a + b =180° …(i)

But a = b +

x 90° = b + 30°

⇒ a – b = 30° …(ii)

Adding (i) and (ii)

210°

2a = 210° ⇒ a =

= 105°

and 105° + b = 180°

⇒ b = 180° – 105°

∴ b = 75°

Hence a = 105°, b = 75°

Question - 18 : -
In the figure, ∠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.

Answer - 18 : -

In the figure,

∠AOF and ∠FOG form a linear pair

∠EOB = ∠FOC = 90°

∠DOC = ∠FOG = ∠AOB = 30°

(i) ∠BOE = 90°, ∠AOB = 30°

But ∠BOE + ∠AOB + ∠EOG = 180°

⇒ 30° + 90° + ∠EOG = 180°

∴∠EOG = 180° – 30° – 90° = 60°

But ∠FOG = 30°

∴ ∠FOE = 60° – 30° = 30°

∠COD = 30°, ∠COF = 90°

∴ ∠DOF = 90° – 30° = 60°

∠DOE = ∠DOF – ∠EOF

= 60° – 30° = 30°

∠BOC = BOE – ∠COE

= 90° – 30° – 30° = 90° – 60° = 30°

(ii) Right angles are,

∠AOD = 30° + 30° + 30° = 90°

∠BOE = 30° + 30° + 30° = 90°

∠COF = 30° + 30° + 30° = 90°

and ∠DOG = 30° + 30° + 30° = 90°

(iii) Pairs of adjacent complementaiy angles are ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠BOC, ∠COE

(iv) Pairs of adjacent supplementary angles are ∠AOB, ∠BOG; ∠AOC, ∠COG and ∠AOD, ∠DOG

(v) Pairs of adjacent angles are ∠BOC, ∠COD; ∠COD, ∠DOE and ∠DOE, ∠EOF.

Question - 19 : - In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = (∠QOS – ∠POS). [NCERT]

Answer - 19 : -

Ray OR ⊥ ROQ. OS is another ray lying between OP and OR.

To prove : ∠ROS =

(∠QOS – ∠POS)

Proof : ∵ RO ⊥ POQ

∴ ∠POR = 90°

⇒ ∠POS + ∠ROS = 90° …(i)

⇒ ∠ROS = 90° – ∠POS

But ∠POS + ∠QOS = 180° (Linear pair)

= 2(∠POS + ∠ROS) [From (i)]

∠POS + ∠QOS = 2∠ROS + 2∠POS

⇒ 2∠ROS = ∠POS + ∠QOS – 2∠POS

= ∠QOS – ∠POS

∴ ROS =

(∠QOS – ∠POS)

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