Chapter 4: Lines and Angles (Exercise 4.1

Chapter 4: Lines and Angles (Exercise 4.1)

Question 1:
In the given figure, AOB is a straight line. Find the value of x, ∠AOC, and ∠BOC.

Solution:
* Since AOB is a straight line, the sum of angles on a straight line is 180° (Linear Pair).
∠AOC + ∠BOC = 180°
(3x + 20)° + (2x - 10)° = 180°
5x + 10 = 180
5x = 170
x = 34°

* Calculating individual angles:
∠AOC = 3(34) + 20 = 102 + 20 = 122°
∠BOC = 2(34) - 10 = 68 - 10 = 58°

Question 2:
In the given figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.

Solution:
* Given: ∠POR : ∠ROQ = 5 : 7
* Let ∠POR = 5a and ∠ROQ = 7a.
* Since POQ is a straight line, ∠POR and ∠ROQ form a linear pair:
∠POR + ∠ROQ = 180°
5a + 7a = 180°
12a = 180°
a = 15°

* Therefore, the angles are:
∠POR = 5 × 15° = 75°
∠ROQ = 7 × 15° = 105°
∠QOS = ∠POR = 75° (Vertically opposite angles)
∠POS = ∠ROQ = 105° (Vertically opposite angles)

Question 3:
In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:
* On straight line ST at point Q:
∠PQS + ∠PQR = 180° --- (Equation 1, Linear pair)

* On straight line ST at point R:
∠PRT + ∠PRQ = 180° --- (Equation 2, Linear pair)

* Equating Equation 1 and Equation 2:
∠PQS + ∠PQR = ∠PRT + ∠PRQ

* Since it is given that ∠PQR = ∠PRQ, subtracting equal terms from both sides:
∠PQS = ∠PRT
(Hence proved.)

Question 4:
In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
* Lines AB and CD intersect at O.
* ∠AOC and ∠BOD are vertically opposite angles:
∠AOC = ∠BOD = 40°

* Given: ∠AOC + ∠BOE = 70°
40° + ∠BOE = 70°
∠BOE = 30°

* Since AOB is a straight line, the sum of angles is 180°:
∠AOC + ∠COE + ∠BOE = 180°
70° + ∠COE = 180°
∠COE = 110°

* Finding Reflex ∠COE:
Reflex ∠COE = 360° - 110° = 250°

Question 5:
In the given figure, if x + y = w + z, then prove that AOB is a straight line.

Solution:
* The sum of all angles around a point is equal to 360°:
x + y + w + z = 360°

* Given that x + y = w + z, substituting w + z with x + y:
(x + y) + (x + y) = 360°
2(x + y) = 360°
x + y = 180°

* Since x and y form a linear pair (∠BOC + ∠AOC = 180°), AOB is a straight line.
(Hence proved.)

Question 6:
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. Ray OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½ (∠QOS - ∠POS).

Solution:
* Since OR ⊥ PQ, we have ∠ROQ = 90° and ∠ROP = 90°.
* ∠ROS = ∠ROP - ∠POS = 90° - ∠POS --- (Equation 1)
* ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS
⇒ ∠ROS = ∠QOS - 90° --- (Equation 2)

* Adding Equation 1 and Equation 2:
2 ∠ROS = (90° - ∠POS) + (∠QOS - 90°)
2 ∠ROS = ∠QOS - ∠POS
∠ROS = ½ (∠QOS - ∠POS)
(Hence proved.)