Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 – Lines and Angles. This Exercise contains 6 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 7 or other Chapters, you can click the link at the end of this Note.
NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 Lines and Angles
NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 Lines and Angles
Q.1 In given figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ
Given :
∠SPR = 135°
∠PQT = 110°
To Find :
∠PRQ = ?
Solution:
∠SPR + ∠RPQ = 180° [ Linear Pair of Angles ]
135° + ∠RPQ = 180°
∠RPQ = 180° – 135°
∠RPQ = 45°
∠PQT + ∠PQR = 180° [ Linear Pair of Angles ]
110° + ∠PQR = 180°
∠PQR = 180° – 110°
∠PQR = 70°
According to Angle Sum Property, In ∆ PQR
∠PQR + ∠PRQ + ∠RPQ = 180°
70° + ∠PRQ + 45° = 180°
∠PRQ + 115° = 180°
∠PRQ = 180° – 115°
∠PRQ = 65°
Q. 2 In given figure, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆ XYZ, find ∠OZY and ∠YOZ
Given :
∠X = 62°
∠XYZ = 54°
YO and ZO are bisectors of ∠XYZ & ∠XZY
To Find :
∠OZY = ?
∠YOZ = ?
Solution:
According to Angle Sum Property, In ∆ XYZ
∠XYZ + ∠YZX + ∠ZXY = 180°
54° + ∠YZX + 62° = 180°
∠YZX + 116° = 180°
∠YZX = 180° – 116°
∠YZX = 64°
∠OZY =  64°/2
∠OZY = 32° [ OZ is angle bisector of ∠YZX ]
∠OYZ =  54°/2 = 27° [ OY is angle bisector of ∠XYZ ]
According to Angle Sum Property, In ∆ OYZ
∠OYZ + ∠OZY + ∠YOZ = 180°
27° + 32° + ∠YOZ = 180°
59° + ∠YOZ = 180°
∠YOZ = 180° – 59°
∠YOZ = 21°
Q.3 In given figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Given :
AB || DE
∠BAC = 35°
∠CDE = 53°
To Find :
∠DCE = ?
Solution:
∠BAC = ∠CED = 35° [ Alternate interior angles ]
According to Angle Sum Property, In ∆ CDE
∠CDE + ∠CED + ∠DCE = 180°
53° + 35° + ∠DCE = 180°
88° + ∠DCE = 180°
∠DCE = 180° – 88°
∠DCE = 92°
Q. 4 In given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Given :
∠PRT = 40°
∠RPT = 95°
∠TSQ = 75°
To Find : ∠SQT = ?
Solution:
According to Angle Sum Property, In ∆ PRT
∠RPT + ∠PRT + ∠PTR = 180°
95° + 40° + ∠PTR = 180°
135° + ∠PTR = 180°
∠PTR = 180° – 135°
∠PTR = 45°
∠PTR = ∠STQ = 45° [ Alternate interior angle ]
According to Angle Sum Property, In ∆ STQ
∠QST + ∠STQ + ∠SQT = 180°
75° + 45° + ∠SQT = 180°
120° + ∠SQT = 180°
∠SQT = 180° – 120°
∠SQT = 60°
Q.5 In given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Given :
PQ ⊥ PS
PQ || SR
∠SQR = 28°
∠QRT = 65°
To Find :
x = ?
y = ?
Solution:
∠PQR = ∠QRT [ Alternate interior angle ]
x + 28° = 65°
x = 65° – 28°
x = 37°
According to Angle Sum Property, In ∆PQS
x + y + ∠QPS = 180°
37° + y + 90° = 180°
y + 127° = 180°
y = 180° – 127° = 53°
Q.6 In given figure, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR
Solution:
In ∆ QTR, ∠TRS is an Exterior Angle.
∠QTR + ∠TQR = ∠TRS
∠QTR = ∠TRS – ∠TQR ——– 1
In ∆ PQR, ∠PRS is an Exterior Angle.
∠QPR + ∠PQR = ∠PRS
∠QPR + 2 ∠TQR = 2 ∠TRS
∠QPR = 2 ( ∠TRS – ∠TQR )
Using Eq 1, ∠QPR = 2 QTR
∠QTR = 1/2 ∠QPR
NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 – Lines and Angles, has been designed by the NCERT to test the knowledge of the student on the topic – Angle Sum Property of a Triangle
The next Exercise for NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 – Triangles can be accessed by clicking here.
Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 – Lines and Angles
