Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.1 Textbook Exercise Questions and Answers.
Haryana Board 9th Class Maths Solutions Chapter 6 Lines and Angles Exercise 6.1
Question 1.
In figure 6.26, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
∠AOC = ∠BOD, (Vertically opposite angles)
⇒ ∠AOC = 40°,
∵ ∠BOD = 40°, given …(i)
Now, ∠AOC + ∠BOE = 70°, (Given)
⇒ 40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30° …(ii)
Ray OC stands on line AB.
∴ ∠AOC + ∠COB = 180°,
(Linear pair axiom)
⇒ ∠AOC + ∠COE + ∠BOE = 180°,
∵ ∠COB = ∠COE + ∠BOE
⇒ 40° + ∠COE +30° = 180°, (From (i) and (ii), we have ∠AOC = 40°, ∠BOE = 30°)
⇒ ∠COE + 70° = 180°
⇒ ∠COE = 180° – 70° = 110°
So, Reflex
∠COE = 360° – 110° = 250°
Hence, ∠BOE = 30°
and reflex ∠COE = 250°.
Question 2.
In figure 6-27, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
Since ray OP stands on line XY. Therefore ∠POY + ∠POX = 180°,
(Linear pair axiom)
⇒ 90° + ∠POX = 180°,
[∵ ∠POY = 90°, given]
⇒ ∠POX = 180° – 90° = 90°
⇒ a + b = 90°
But a : b= 2 : 3 (given)
Sum of ratios = 2 + 3 = 5
∴ a = \(\frac {2}{5}\) × 90° = 36°
and b = \(\frac {3}{5}\) × 90° = 54°
Now, MN is a line.
Since, ray OX stands on line MN.
Therefore, ∠MOX + ∠XON = 180°,
(Linear pair axiom)
⇒ b + c = 180°
⇒ 54° + c = 180°
⇒ c = 180° – 54° = 126°
Hence, c = 126°.
Question 3.
In figure 6.28, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution :
Since, ray QP stands on line SQT. Therefore, ∠PQS + ∠PQR = 180°,
(Linear pair axiom) …(i)
Since, ray RP stands on line SQT. Therefore, ∠PRT + ∠PRQ = 180°,
(Linear pair axiom) ..(ii)
From (i) and (ii), we get
∠PQS + ∠PQR = ∠PRT + ∠PRQ
⇒ ∠PQS + ∠PQR = ∠PRT + ∠PQR,
[∵ ∠PRQ = ∠PQR, given)
⇒ ∠PQS = ∠PRT, [Subtracting ∠PQR from both sides]
Hence proved
Question 4.
In figure 6.29, if x + y = w + z, then prove that AOB is a line.
Solution:
Since, sum of all the angles round a point is equal to 360°. Therefore,
∠AOC + ∠BOC + ∠BOD + ∠AOD = 360°
⇒ y + x + w + z = 360°
⇒ x + y + x + y = 360° [Given w + z = x + y]
⇒ 2x + 2y = 360°
⇒ 2(x + y) = 360°
⇒ x + y = \(\frac {360°}{2}\) = 180°
According to Axiom 6.2, if the sum of two adjacent angles is 180°, then non-common arms of the angles form a line.
Hence, AOB is a line. Hence Proved
Question 5.
In figure 6.30, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that:
∠ROS = \(\frac {1}{2}\)(∠QOS – ∠POS).
Solution:
Since, ray OR ⊥ line PQ.
∠ROQ + ∠ROP = 180°,
(Linear pair axiom)
⇒ 90° + ∠ROP = 180° [∵∠ROQ = 90°)
⇒ ∠ROP = 180° – 90° = 90°
⇒ ∠ROS + ∠POS = 90°,
∵ [∠ROP = ∠ROS + ∠POS)
⇒ ∠POS = 90° – ∠ROS …(i)
∠QOS =∠ROS + ∠ROQ
⇒ ∠QOS = ∠ROS + 90° …(ii)
Subtracting (i) from (ii), we get
∠QOS – ∠POS = ∠ROS + 90° – (90° – ∠ROS)
⇒ ∠QOS – ∠POS = ∠ROS + 90° – 90° + ∠ROS
⇒ ∠QOS – ∠POS = 2∠ROS
⇒ \(\frac {1}{2}\)(∠QOS – ∠POS) = ∠ROS
⇒ ∠ROS = \(\frac {1}{2}\)(∠QOS – ∠POS).
Hence proved
Question 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.
Solution:
It is given that XY produced to P. So, XP is a straight line.
Now, ray YZ stands on line XP.
∴ ∠XYZ + ∠ZYP = 180°
(Linear pair axiom)
⇒ 64° + ∠ZYP = 180°, [∵∠XYZ = 64°, given]
⇒ ∠ZYP = 180° – 64° = 116°
Since, ray YQ bisects ∠ZYP.
∠ZYQ = ∠QYP = \(\frac {116°}{2}\) = 58°
Now, ∠XYQ = ∠XYZ + ∠ZYQ
= 64° + 58°
= 122°
Reflex ∠QYP = 360° – ∠QYP
= 360° – 58° = 302°
Hence, ∠XYQ = 122° and Reflex ∠QYP = 302°.