Lines and Angles NCERT Solution, study material, CBSE Notes

NCERT Solution: Lines and Angles

In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer
Given,
∠AOC + ∠BOE = 70° and ∠BOD = 40°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°

In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Given,
∠POY = 90° and a : b = 2 : 3
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°

In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° - ∠PQR --- (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° - ∠PRQ
⇒ ∠PRQ = 180° - ∠PQR --- (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° - ∠PQR
Therefore, ∠PQS = ∠PRT

In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.

In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS --- (i)
∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)
Subtracting (ii) from (i)
∠QOS - ∠POS = 90° + ∠ROS - (90° - ∠ROS)
⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS
⇒ ∠QOS - ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.

It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Given,
∠XYZ = 64°
YQ bisects ∠ZYP

∠XYZ +∠ZYP = 180° (Linear Pair)
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠ZYP = 2∠ZYQ
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 58° = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + 58°
⇒ ∠XYQ = 122°
also,
reflex ∠QYP = 180° + ∠XYQ
∠QYP = 180° + 122°
⇒ ∠QYP = 302°

In Fig. 6.28, find the values of x and y and then show that AB || CD.

x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.

In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Given,
AB || CD and CD || EF
y : z = 3 : 7
Now,
x + y = 180° (Angles on the same side of transversal.)
also,
∠O = z (Corresponding angles)
and, y + ∠O = 180° (Linear pair)
⇒ y + z = 180°
A/q,
y = 3w and z = 7w
3w + 7w = 180°
⇒ 10 w = 180°
⇒ w = 18°
∴ y = 3×18° = 54°
and, z = 7×18° = 126°
Now,
x + y = 180°
⇒ x + 54° = 180°
⇒ x = 126°