**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.**

**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**