Lines and Angles-NCERT Solutions
Class IX Science
NCERT Solutions For Lines and Angles
NCERT Solutions For Lines and Angles
Exercise 6.1 (Page 96)
1. In the following figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Ans. Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180°
or 70° + ∠COE = 180°
[∴ ∠AOC + ∠BOC = 70° (Given)]
or ∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
∴ AB and CD intersect at O.
∴ ∠COA = ∠BOD
[Vertically opposite angles]
But ∠BOD = 40°[Given]
∴ ∠COA = 40°
Also ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70°
or ∠BOE = 70° – 40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250.
2. In the following figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Ans. XOY is a straight line.
∴ ∠b + ∠a + ∠POY = 180°
But ∠OPY = 90° [Given]
∴ ∠b + ∠a = 180° – 90° = 90°.
Also a : b = 2 : 3
Since XY and MN intersect at O,
∴ ∠c = [∠a + ∠POY]
[Vertically opposite angles]
or ∠c = 36° + 90° = 126°
Thus, the required measure of ‘c’
3. In the following figure, ∠PAR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Ans. ∴ ST is a straight line,
∴ ∠PQS + ∠PAR = 180° ...(1)
Similarly, ∠PRT + ∠PRQ = 180° ...(2)
From (1) and (2), we have
∠PQS + ∠PQR = ∠PRT + ∠PRQ
But ∠PQR = ∠PRQ [Given]
∴ ∠PQS = ∠PRT
4. In the following figure, if x + y = w + z, then prove that AOB is a line.
Ans. ∴ Sum of all the angles at a point = 360°
∴ x + y + z + w = 360°
or (x + y) + (z + w) = 360°
But (x + y) = (z + w) [Given]
∴ (x + y) + (x + y) = 360°
or 2(x + y) = 360°
∴ AOB is a straight line.
5. In the adjoining figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that 
Ans. ∴ POQ is a straight line. [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180°
But OR ⊥ PQ
∴ ∠ROQ = 90°
∴ ∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 90° ...(1)
Now, we have ∠ROS + ∠ROQ = ∠QOS
⇒ ∠ROS + 90° = ∠QOS ...(2)
From (1) and (2), we have
∠ROS + [∠POS + ∠ROS] = ∠QOS
⇒ 2∠ROS + ∠POS = ∠QOS
⇒ 2∠ROS = [∠QOS – ∠POS]
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.
Ans. ∴ XYP is a straight line.
∠XYZ + ∠ZYQ + ∠QYP = 180°
⇒ 64° + ∠ZYQ + ∠QYP = 180°
[∴ YQ, bisects ∠ZYP, so ∠QYP = ∠ZYQ]
⇒ 64° + 2∠QYP = 180°
⇒ 2∠QYP = 180° – 64° = 116°
∴ Reflex ∠QYP = 360° – 58° = 302°
Since ∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP
[∴ ∠XYZ = 64° (Given) and ∠ZYQ = ∠ QYP]
[∴ ∠QYP = 58°]
⇒ ∠XYQ = 64° + 58°
= 122°
Thus, ∠XYQ = 122° and reflex ∠QYP = 302°
Exercise 6.2 (Page 103)
1. In the following figure, find the values of x and y and then show that AB || CD.
Ans. In the figure, we have CD and PQ intersect at Y.
∴ y = 130° [Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∴ ∠AEP + ∠AEQ = 180° [Linear pair]
or 50° + x = 180°
⇒ x = 180° – 50° = 130°
...(2)
From (1) and (2), x = y
But they are the angles of a pair of interior alternate angles.
∴ AB || CD.
2. In the following Figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Ans. 
∴ AB|| EF and PQ is a transversal.
∴ Interior alternate angles are equal.
∴ ∠x = ∠z
...(1)
Again, AB || CD,
∴ Interior opposite angles are supplementary.
⇒ y + z = 180°
But y : z = 3 : 7
From (1) and (2), we have
x = 126°
3. In the following figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Ans. AB || CD and GE is a transversal.
∴ Interior alternate angles are equal.
∴ ∠AGE = ∠GED
But ∠GED = 126°
[Given]
∴ ∠AGE = 126°
Since ∠GED = 126°
∴ ∠GEF + ∠FED = ∠GED
or ∠GEF + 90° = 126°
or ∠GEF = 126 – 90° = 36°
Next, AB || CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180°
or ∠FGE + 126° = 180°
or ∠FGE = 180° – 126° = 54°
Thus, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°
4. In the following figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
Hint: Draw a line parallel to ST through point R.
Ans. ∴ PQ || ST
[Given]
and EF || ST
[Construction]
∴ PQ || EF
and QR is a transversal,
∴ Interior alternate angles are equal i.e ∠PAR = ∠QRF
But ∠PAR = 110°
[Given]
∴ ∠QRF = ∠QRS + ∠SRF = 110°
...(1)
Again ST || EF [Construction] and RS is a transversal.
∴ ∠RST + ∠SRF = 180°
or 130° + ∠SRF = 180°
⇒ ∠SRF = 180° – 130° = 50°
Now, from (1), we have
∠QRS + 50° = 110°
⇒ ∠QRS = 110° – 50° = 60°
Thus, ∠QRS = 60°.
5. In the following figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Ans. We have AB || CD
[Given]
and PQ is a transversal.
∴ Interior alternate angles are equal.
∴ ∠APQ = ∠PQR
or 50° = x[∴ APQ = 50° (Given)]
...(1)
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD
[Interior alternate angles]
⇒ ∠APR = 127°
[∴ It is given that ∠PRD = 127°]
But ∠APR = ∠APQ + ∠QPR
∴ ∠APQ + ∠QPR + 127°
⇒ 50° + y = 127°
[∴ It is given that ∠APQ = 50°]
⇒ y = 127° – 50° = 77°
Thus, x = 50° and y = 77°
6. In the following figure, PQ and RS are two mirrors placed parallel to each other, An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
Ans.
Remember
(i) Perpendiculars to the parallel lines are parallel.
(ii) According to the laws of reflection, angle of incidence = angle of reflection
Draw ray BL ⊥ PQ and CM ⊥ RS
∵ PQ || RS
[Given]
∴ BL || CM [∴ BL ⊥ PQ and CM ⊥ RS]
and BC is a transversal.
∴ ∠LBC = ∠MCB
[Interior alternate angles]
Since,
(Angle of incidence) = (Angle of reflection)
∴ ∠ABL = ∠LBC and ∠MCD = ∠MCD
⇒ ∠ABL = ∠MCD
∴ From (1), we have
∠LBC + ∠ABL = ∠MCD + ∠MCD
[Equals are added to equals]
⇒ ∠ABC = ∠BCD
i.e. Angles of a pair of interior alternate angles are equal,
∴ AB || CD
Exercise 6.3 (Page 107)
1. In the adjoining figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Ans. ∵ TQR is a straight line,
∴ ∠TQP + ∠PQR = 180°
[Linear pair]
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110° = 70°
Since, the side QP of ΔPQR is produced to S.
∴ Exterior angles so formed is equal to the sum of interior opposite angles.
∴ ∠PQR + ∠PRQ = 135°
⇒ 70° + ∠PRQ = 135°
[∴ ∠PQR = 70°]
⇒ ∠PRQ = 135° – 70°
⇒ ∠PRQ = 65°
2. In the adjoining figure, ∠X = 65°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.
Ans. In ∠XYZ,
∠XYZ + ∠YZX + ∠ZXY = 180°
[∵ Sum of angles of a triangle is 180°]
But ∠XYZ = 54° and ∠ZXY = 62°
[Given]
∴ 54° + ∠YZX + 62° = 180°
⇒ ∠YZX = 180° – 54° – 62° = 64°
∴ YO and ZO are the bisectors ∠XYZ and ∠XZY respectively,
[Given]
Now, in ΔOYZ, we have:
∠YOZ + ∠OYZ + ∠OZY = 180°
[By the angle sum property]
⇒∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° – 27° – 32° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121°
3. In the following figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Ans. ∴ AB || DE and AE is a transversal.
[Given]
∴ ∠BAC = ∠AED [Interior alternate angles]
But ∠BAC = 35°
∴ ∠AED = 35°
Now, in ΔCDE, we have
∠CDE + ∠DEC + ∠DCE = 180°
[Using the angle sum property]
∴ 53° + 35° + ∠DCE = 180°
[∴ ∠DEC = ∠AED = 35° and ∠CDE = 53 (Given)]
⇒ ∠DCE = 180° – 53° – 35° = 92°
Thus, DCE = 92°
4. In the figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Ans. In ΔPRT
∠P + ∠R + ∠PTR = 180°
[By the angle sum property]
⇒ 95° + 40° + ∠PTR = 180°
⇒ ∠PTR = 180° – 95° – 40° = 45°
But PQ and RS intersect at T,
∴ ∠PTR = ∠QTS
[Vertically opposite angles]
∴ ∠QTS = 45°
Now, in ΔTQS, we have
∠TSQ + ∠STQ + ∠SQT = 180°
[By angle sum property]
∴ 75° + 45° + ∠SQT = 180°
[∴ ∠TSQ = 75° and ∠STQ = 45°]
⇒ ∠SQT = 180° – 75° – 45° = 60°
Thus, ∠SQT = 60°
5. In the adjoining figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Ans. In ΔQRS, the side SR is produced to T.
∴ Exterior ∠QRT = ∠RQS + ∠RSQ
But ∠RQS = 28° and ∠QRT = 65°
∴ From ∠RQS + ∠RSQ = ∠QRT, we have
28° + ∠RSQ = 65°
⇒ ∠RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
[Given]
∴ ∠PQS = ∠RSQ
[Interior alternate angles]
⇒ x = 37°
Again, PQ ⊥ PS
[Given]
∴ ∠P = 90°
Now, in ΔPQS, we have
∠P + ∠PQS + ∠PSQ = 180°
[By angle sum property]
⇒ 90° + x + y = 180°
⇒ 90° + 37° + y = 180°
[∴ x = 37°]
⇒ y = 180° – 90° – 37° = 53°
Thus, x = 37° and y = 53°
6. In the adjoining figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that 
Ans. In ΔPQR, the side QR is produced to S.
∴ Exterior ∠PRS = Sum of the interior opposite angles
⇒ ∠PRS = ∠P + ∠Q
Since QT and RT are bisectors of ∠Q and ∠PRS respectively,
Now, In ΔQRT, we have
Exterior ∠TRS = ∠TQR + ∠T
...(2)
From (1) and (2), we have
