NCERT Solutions Class 9 Maths
Chapter – 6 (Lines And Angles)
The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 6 Lines And Angles Exercise 6.1 has been provided here to help the students in solving the questions from this exercise.
Chapter 6: Lines and Angles
Exercise – 6.1
1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Answer – From the diagram, we have
∠AOC + ∠BOE +∠COE and ∠COE +∠BOD + ∠BOE form a straight line.
So, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180o
Now, by putting the values of ∠AOC + ∠BOE = 70o and ∠BOD = 40o we get,
70o +∠COE = 180o
∠COE = 110° and ∠BOE = 30°
So, reflex ∠COE = 360o – 110o = 250o
2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Answer – Line OP is perpendicular to line XY. Hence ∠POY = ∠POX = 90°
∠POX = ∠POM + ∠MOX
90° = a + b —————(1)
Since a and b are in the ratio 2 : 3 that is,
a = 2x and b = 3x —————(2)
Substituting (2) in (1),
a + b = 90°
2x + 3x = 90°
5x = 90°
x = 90°/5 = 18°
a = 2x = 2 × 18°
a = 36°
b = 3x = 3 × 18°
b = 54°
Also , ∠MOY= ∠MOP + ∠POY
= a + 90°
= 36° + 90° = 126°
Lines MN and XY intersect at point O and the vertically opposite angles formed are equal.
∠XON = ∠MOY
c = 126°
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Answer – Since ST is a straight line so,
∠PQS + ∠PQR = 180° (linear pair) and
∠PRT + ∠PRQ = 180° (linear pair)
Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°
Since ∠PQR =∠PRQ (as given in the question)
∠PQS = ∠PRT. (Hence proved).
4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
Answer – For proving AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x + y = 180°
We know that the angles around a point are 360° so,
x + y + w + z = 360°
In the question, it is given that,
x + y = w + z
So, (x+y) + (x+y) = 360°
2 (x+y) = 360°
∴ (x+y) = 180° (Hence proved).
5. 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 = ½ (∠QOS – ∠POS).
Answer – In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°
We can write it as ∠ROP = ∠ROQ = 900
We know that
∠ROP = ∠ROQ
It can be written as
∠POS + ∠ROS = ∠ROQ
∠POS + ∠ROS = ∠QOS – ∠ROS
∠SOR + ∠ROS = ∠QOS – ∠POS
So we get
2∠ROS = ∠QOS – ∠POS
Or, ∠ROS = (∠QOS – ∠POS) (Hence proved).
6. 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.
Answer – Here, XP is a straight line
So, ∠XYZ + ∠ZYP = 180°
Putting the value of ∠XYZ = 64° we get,
64° +∠ZYP = 180°
∴ ∠ZYP = 116°
From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP
Now, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
∴ ∠ZYQ = ∠QYP = 58°
Again, ∠XYQ = ∠XYZ + ∠ZYQ
By putting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.
∠XYQ = 64°+58°
Or, ∠XYQ = 122°
Now, reflex ∠QYP = 180°+XYQ
We computed that the value of ∠XYQ = 122°.
So,
∠QYP = 180°+122°
∴ ∠QYP = 302°
