9th Maths

Lines Angles

Exercise 6.1

Question 1: In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°; find ∠BOE and reflex ∠COE.

lines and angles 2

Solution: Given: ∠AOC + ∠BOE = 70° ……………equation (i)
And, ∠BOD = 40°
Now, ∠AOC = ∠BOD (Vertically opposite angles)
Hence, ∠AOC = 40° …………equation (ii)

Now, putting the value of equation (ii) in equation (i);
∠AOC + ∠BOE = 70°
Or, 40° + ∠BOE = 70°
Or, ∠BOE = 70° - 40°= 30°

Now, ∠AOC + ∠BOE + ∠COE = 180° (Angles at a common point on a line)
Or, 70° + ∠COE = 180°
Or, ∠COE = 180° - 70° = 110°
Hence, Reflex ∠COE = 360° - 110° = 250°
Hence, ∠BOE = 30° and reflex ∠COE = 250°




Question 2: In the following figure, lines XY and MN intersect at O. If ∠POY = 90°and a:b = 2:3, find c.

lines and angles 5

Solution: Given: ∠POY = 90°
And a:b = 2:3
Or, a/b = 2/3
Or, a = 2b/3 ………equation (i)

Now, ∠POX + ∠POY = 180°
Or, ∠POX + 90° = 180°
Or, ∠POX = 180° - 90° = 90°
Or, a + b = 90° (because ∠POX = a + b)
Or, 2b/3 + b = 90°
Or, (2b + 3b)/3 = 90°
Or, 5b = 270°
Or, b = 270°/5 = 54°

Putting the value of b in equation (i)
a = 2/3 b
Or, a = 2/3 x 54º
Or, a = 2 x 18º
Or, a = 36º

Now, b + c = 180º {Angles at a common point on a line}
Or, 54º + c = 180º
Or, c = 180º- 54º
Or, c = 126º
So, c = 126º.


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

lines and angles 8

Solution: Given: ∠PQR = ∠PRQ
To Prove: ∠PQS = ∠PRT

Proof: ∠PQR + ∠PQS = 180° (Linear pair of angles)
Similarly, ∠PRQ + ∠PRT = 180°

From above equations;
∠PQR + ∠PQS = ∠PRQ + ∠PRT
Or, ∠PQR + ∠PQS = ∠PQR + ∠PRT (Because ∠PQR = ∠PRQ given)
Or, ∠PQR + vPQS - ∠PQR = ∠PRT
Or, ∠PQS = ∠PRT Proved

Question 4: In the figure if x + y = w + z, then prove that AOB is a line.

lines and angles 10

Given: x + y = w + z

To Prove: AOB is a straight line

Proof:
x + y = w + z ………equation (i) {given}
But,x + y + w + z = 360º {Angles around a point}
Or, (x + y) + (w + z) = 360º
Or, (x + y) + (x + y) = 360º {from equation (i)}
Or, x + y + x + y = 360º
Or, 2x + 2y = 360º
Or, 2(x + y) = 360º
Or, x + y = 360°/2
Or, x + y = 180º
Therefore, x and y form a linear pair.
Hence, AOB is a straight line. Proved.


Question5: In the given 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 = ½(∠QOS - ∠POS)

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Solution: Given: POQ is a straight line.
OR ┴ PQ
Ray OS meets line PQ at O.

Proof: ∠QOS = ∠ROS + ∠ROQ …………..equation (i)
∠POS = ∠POR - ∠ROS ----------equation (ii)

Subtracting equation (ii) from equation (i),
∠QOS - ∠POS = (∠ROS + ∠ROQ) – (∠POR - ∠ROS)
Or, ∠QOS - ∠POS = ∠ROS + ∠ROQ - ∠POR + ∠ROS
Or, ∠QOS - ∠POS = 2∠ROS + ∠ROQ - ∠POR
Or, ∠QOS - ∠POS = 2∠ROS + 90° - 90° (Because OR makes right angle with PQ)
Or, ∠QOS - ∠POS = 2∠ROS
Or, ½(∠QOS - ∠POS) = ∠ROS
Or, ∠ROS = ½(∠QOS - ∠POS) Proved

Question 6: It is given that ∠XYZ = 64° and XY is produced to a point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

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Given: ∠XYZ = 64°
Now, ∠XYZ + ∠ZYP = 180°(Linear pair)
Or, 64° + ∠ZYP = 180°
Or, ∠ZYP = 180° - 64° = 116°

Since, YQ bisects ∠ZYP
So, ∠ZYQ = ∠PYQ = ½ ∠ZYP = 116°/2 = 58°
So, ∠XYQ = ∠XYZ + ∠ZYQ = 64°+ 58° = 122°
Now, reflex ∠QYP = 360° - 58° = 302°