NCERT Solutions class 9 maths exercise 6.1-Lines and Angles
NCERT solutions class 9 maths exercise 6.1 of chapter 6-Lines and Angles is an easy chapter since the students of 9 class already would have studied Lines and Angles in their previous classes. In this chapter 6- Lines and Angles, the questions are based on lines and angles. All unsolved questions of exercise 6.1 are solved by an expert of maths as per the CBSE norms by a step by step method with a suitable diagram.
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NCERT Solutions class 9 maths Chapter 6-Lines and Angles
NCERT Solutions of class 9 maths
Q1.In fig.6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Ans. We are given
∠AOC + ∠BOE = 70°
∠BOD = 40°
∠BOD = ∠AOC (Vertically opposite angle)
∴ ∠AOC = 40°
Putting values of ∠AOC in the given equation ∠AOC + ∠BOE = 70°
40 + ∠BOE = 70
∠BOE = 30°
∠AOE + ∠BOE = 180° (Linear pair)
∠AOC + ∠COE + ∠BOE = 180°
40° + ∠COE + 30° = 180°
∠COE = 180° – 70° = 110°
Reflex ∠COE = 360° – ∠COE
Reflex ∠COE = 360° – 110° =250°
Therefore ∠BOE =30° and reflex ∠COE =250°
Q2.In fig.6.4 ,lines XY and MN intersect at O .If ∠POY =90° and a: b =2: 3,find c.
Ans.We are given
∠POY =90° and a: b =2: 3
∠POX = 180° – 90° = 90°(Linear pair)
Let a = 2x and b = 3x
∠POX = a + b = 90
2x + 3x = 90
5x = 90°
x = 18°
Therefore a = 2×18 = 36° and b = 3× 18 = 54°
Since MON is a line therefore b and c are linear pair
∴b + c =180°
54° + c = 180°
c = 180° – 54° = 126°
Hence the value of c is 126°
Q3. In fig. 6.15 ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT
Ans.
Given: ∠PQR = ∠PRQ
To Prove: ∠PQS = ∠PRT
Proof: In the given fig.
∠PQS + ∠PQR = 180°(Linear pair)…..(i)
∠PRT + ∠PRQ = 180°(Linear pair)…..(ii)
From equations (i) and (ii)
∠PQS + ∠PQR = ∠PRT + ∠PRQ
∠PQR = ∠PRQ (given)
∠PQS + ∠PRQ = ∠PRT + ∠PRQ
∠PQS = ∠PRT
Hence Proved
Q4. In fig.6.16 if x +y = w +z, then prove that AOB is a line.
Ans. We have In the given figure
Given: x +y = w +z
To Prove: AOB is a line
Proof: From the figure, we can observe that all the given angles form a complete angle of 360°
Therefore we can write
x +y + w +z = 360° (complete angle)
w + z + w + z = 360°
2w + 2z = 360°
2(w + z) = 360°
w + z = 180
The sum of w and z implies that both angle w and z forms a linear pair, so AOB is a line.
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Q5. 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).
Ans.
Given: POQ is a line
OR ⊥ PQ
∴ ∠POR = 90° and ∠QOR = 90°
To Prove: ∠ROS = 1/2(∠QOS -∠POS)
Proof: POQ is a line (given)
Therefore ∠POR and ∠QOR are the linear pair
∠POR + ∠QOR = 180°
∠POS + ∠ROS + 90° = 180°
∠POS + ∠ROS = 90°…..(i)
From the fig. we have
∠QOS = ∠QOR + ∠ROS
∠QOS = 90° + ∠ROS
∠QOS – ∠ROS = 90°……(ii)
From (i) and (ii)
∠POS + ∠ROS = ∠QOS – ∠ROS
∠ROS + ∠ROS = ∠QOS – ∠POS
2∠ROS = ∠QOS – ∠POS
∠ROS = 1/2(∠QOS – ∠POS)
Hence proved
Q6.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.
Ans.
We are given
∠XYZ = 64°
Let ∠PYZ = 2x
∠QYZ = ∠QYP = x ( YQ is the bisector of ∠PYZ)
Since PYX is a line,so
∠PYZ +∠XYZ = 180°
∠QYZ + ∠QYP + 64° = 180°
x + x = 180° – 64° = 116°
2x = 116°
x = 58°,so ∠QYZ = ∠QYP = 58°
∴∠XYQ = ∠XYZ + ∠QYZ = 64° + 58° = 122°
Since ∠QYP = 58°, therefore reflex ∠QYP = 360°-58° = 302°
Hence ∠XYQ = 122° and reflex ∠QYP = 302°
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