**NCERT Solutions Class 9 Maths exercise 6.3 -Lines and Angles**

**NCERT Solutions Class 9 Maths exercise 6.3 of the chapter 6-Lines and Angles** are the **solutions** of the unsolved questions of** exercise 6.3** **-Lines and angles of class 9** **maths**. The questions of **exercise 6.3** are easy to understand because the concept of geometry is based on the **maths** of previous **classes** as an example angle sum property of the triangles, the relationship between exterior angle and its opposite interior angles of a triangle, linear pair etc. Here you can study NCERT solutions of science and maths from class 9 to class 12, our articles on science and maths, sample papers, previous year question papers, tips for government and other entrance exams, and career in online jobs.

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**NCERT Solutions of class 9 maths**

**Q1.In fig. 6.39, side QP and RQ of ΔPQR are produced to points S and T respectively, if ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.**

Ans.We are given in figure ∠SPR = 135° and ∠PQT = 110°

∠PQT and ∠PQR are the linear pair(TQR is a line)

So, ∠PQT + ∠PQR = 180°

110°+ ∠PQR = 180°

∠PQR = 180°- 110° = 70°

∠QPR and ∠SPR are the linear pair (QPS is a line)

∠QPR + ∠SPR = 180°

∠QPR + 135° = 180°

∠QPR = 180° – 135° = 45°

∠PQR +∠QPR +∠PRQ = 180°(angle sum property of triangle)

∠PRQ + 70° + 45° = 180°

∠PRQ + 115° = 180°

∠PRQ = 180° – 115° = 65°

**Q2. In fig.6.40 , ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.**

Ans.We are given ∠X = 62°, ∠XYZ = 54°

YO is the bisector of ∠XYZ and ZO is the bisector of ∠XZY

∴∠OYZ = 54°/2 = 27° and ∠OZY = ∠XZY/2

∠X + ∠XYZ + ∠XZY = 180° (angle sum property of triangle)

62° + 54° + ∠XZY = 180°

116 + ∠XZY = 180°

∠XZY = 180° – 116 = 64°

Now ,In ΔYOZ we have

∠OZY = ∠XZY/2 = 64°/2 = 32°

Applying the angle sum property of triangle in ΔYOZ

∠OYZ + ∠OZY + ∠YOZ = 180°

27° + 32° + ∠YOZ = 180°

59 + ∠YOZ = 180°

∠YOZ = 180° – 59 = 121°

Hence the value of ∠OZY is 32° and of ∠YOZ= 121°

**Q3.In fig.6.41 ,if AB ll DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.**

Ans.We are given AB ll DE, ∠BAC = 35° and ∠CDE = 53°

Since AB ll DE and AE is transversal

So, ∠BAC = ∠DEC (alternate angle)

∠DEC = 35°

Applying the angle sum property of triangle in ΔDEC

∠DEC + ∠DCE + ∠CDE = 180°

35° + 53° + ∠DCE = 180°

88 + ∠DCE = 180°

∠DCE = 180° – 88 = 92°

Hence the value of the ∠DCE is 92°

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**Q4.In fig.6.42 ,if lines PQ and RS intersect at point T, such that ∠PRT =40°,∠RPT =95° and ∠TSQ = 75°, find ∠SQT.**

Ans. We are given in the fig.

∠PRT =40°,∠RPT =95° and ∠TSQ = 75°

Applying the angle sum property of the triangle in ΔPRT

∠RPT +∠PRT+ ∠PTR = 180°

95° + 40° + ∠PTR = 180°

135° + ∠PTR = 180°

∠PTR = 180°- 135° = 45°

Since PQ and RS are the lines

So, ∠PTR = ∠QTS (Vertically opposite angle)

∠QTS = 45°

Applying the angle sum property of triangle in ΔQTS

∠QTS + ∠TSQ + ∠SQT = 180°

45° + 75° + ∠SQT = 180°

120° + ∠SQT = 180°

∠SQT = 180°- 120° = 60°

Hence the value of ∠SQT is 60°

**Q5. In fig.6.43, if PQ ⊥ PS, PQ ll SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.**

Ans.In the fig. we are given PQ ⊥ PS ⇒∠QPS =90°, PQ ll SR, ∠SQR = 28° and ∠QRT = 65°

Since PQ ll SR and QR is the transversal,so∠PQR (x + 28) and ∠QRT(65°) are the alternate angles.

x + 28° = 65°

x = 65° – 28° = 37°

Applying the angle sum property of triangle in ΔQPS

∠QPS +y+ x = 180°

90° + y + x = 180°

x + y = 180° – 90° = 90°

Putting the value of x=37°

y + 37° = 90°

y = 90°- 37° = 53°

Therefore the value of x is 37° and of y is 53°

**Q6. In fig.6.44 ,the side QR of ΔPQR is produced to a point S.If the bisector of ∠PQR and ∠PRS meet at point T,then prove that ∠QTR = 1/2 ∠QPR.**

Ans. In the fig. we are given QT is the bisector of ∠PQR and TR is the bisector of ∠PRS

∴∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR

In ΔPQR , ∠PRS is the exterior angles of ΔPQR and ∠PQR and ∠QPR are the opposite interior angles.

We know the exterior angle of a triangle is the sum of opposite interior angles.

∠PRS = ∠QPR + ∠PQR……..(i)

In ΔQTR, ∠TRS is the exterior angles of ΔQTR and ∠TQR and ∠QTR are the opposite interior angles.

We know the exterior angle of a triangle is the sum of opposite interior angles

∠TRS = ∠TQR + ∠QTR

Since, we are given that ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR

∠PRS/2 = ∠PQR/2 + ∠QTR

∠PRS = ∠PQR + 2∠QTR…….(ii)

From the equation (i) and (ii), we have

∠QPR + ∠PQR = ∠PQR + 2∠QTR

∠QTR = 1/2 ∠QPR, Hence proved∴

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