NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.3

NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles

Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.3 In This Chapter We Will Learn About Lines and Angles, Line-segment, Collinear points, Angle, Arms, Vertex, Ray, Acute, Right angle, Obtuse angle, Straight angle, Reflex angle, Complementary angle, Supplementary angles, Adjacent, Vertically opposite angle,  with Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.3.

NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles

Chapter -6

Lines and Angles

Exercise – 6.3

Question 1. In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ. Solution: We have, ∠TQP + ∠PQR = 180° [Linear pair] ⇒ 110° + ∠PQR = 180° ⇒ ∠PQR = 180° – 110° = 70° Since, the side QP of ∆PQR is produced to S. ⇒ ∠PQR + ∠PRQ = 135° [Exterior angle property of a triangle] ⇒ 70° + ∠PRQ = 135° [∠PQR = 70°] ⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65°

Question 2. In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ. Solution: In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180° [Angle sum property of a triangle] But ∠XYZ = 54° and ∠ZXY = 62° ∴ 54° + ∠YZX + 62° = 180° ⇒ ∠YZX = 180° – 54° – 62° = 64° YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively. ∴ ∠OYZ = 12∠XYZ = 12(54°) = 27° and ∠OZY = 12∠YZX = 12(64°) = 32° Now, in ∆OYZ, we have ∠YOZ + ∠OYZ + ∠OZY = 180° [Angle sum property of a triangle] ⇒ ∠YOZ + 27° + 32° = 180° ⇒ ∠YOZ = 180° -27° – 32° = 121° Thus, ∠OZY = 32° and ∠YOZ = 121°

Question 3. In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE. Solution: AB || DE and AE is a transversal. So, ∠BAC = ∠AED [Alternate interior angles] and ∠BAC = 35° [Given] ∴ ∠AED = 35° Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180° [Angle sum property of a triangle] ∴ 53° + 35° + ∠DCE =180° [∵ ∠DEC = ∠AED = 35° and ∠CDE = 53° (Given)] ⇒ ∠DCE = 180° – 53° – 35° = 92° Thus, ∠DCE = 92°

Question 4. In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT. Solution: In ∆PRT, we have ∠P + ∠R + ∠PTR = 180° [Angle sum property of a triangle] ⇒ 95° + 40° + ∠PTR =180° [ ∵ ∠P = 95°, ∠R = 40° (given)] ⇒ ∠PTR = 180° – 95° – 40° = 45° But PQ and RS intersect at T. ∴ ∠PTR = ∠QTS [Vertically opposite angles] ∴ ∠QTS = 45° [ ∵ ∠PTR = 45°] Now, in ∆ TQS, we have ∠TSQ + ∠STQ + ∠SQT = 180° [Angle sum property of a triangle] ∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°] ⇒ ∠SQT = 180° – 75° – 45° = 60° Thus, ∠SQT = 60°

Question 5. In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y. Solution: In ∆ QRS, the side SR is produced to T. ∴ ∠QRT = ∠RQS + ∠RSQ [Exterior angle property of a triangle] But ∠RQS = 28° and ∠QRT = 65° So, 28° + ∠RSQ = 65° ⇒ ∠RSQ = 65° – 28° = 37° Since, PQ || SR and QS is a transversal. ∴ ∠PQS = ∠RSQ = 37° [Alternate interior angles] ⇒ x = 37° Again, PQ ⊥ PS ⇒ AP = 90° Now, in ∆PQS, we have ∠P + ∠PQS + ∠PSQ = 180° [Angle sum property of a triangle] ⇒ 90° + 37° + y = 180° ⇒ y = 180° – 90° – 37° = 53° Thus, x = 37° and y = 53°

• Exercise – 6.1
• Exercise – 6.2
  

NCERT Solutions Class 9th Maths All Chapter

• Chapter 1 – Number systems
• Chapter 2 – Polynomials
• Chapter 3 – Coordinate Geometry
• Chapter 4 – Linear Equations in Two Variables
• Chapter 5 – Introduction to Euclid Geometry
• Chapter 6 – Lines and Angles
• Chapter 7 – Triangles
• Chapter 8 – Quadrilaterals
• Chapter 9 – Areas of Parallelograms and Triangles
• Chapter 10 – Circles
• Chapter 11 – Constructions
• Chapter 12 – Heron’s Formula
• Chapter 13 – Surface Areas and Volumes
• Chapter 14 – Statistics
• Chapter 15 – Probability