 NCERT Solutions class 9 Maths exercise 6.3 of chapter 6-Lines and Angles - Future Study Point # NCERT Solutions class 9 Maths exercise 6.3 of chapter 6-Lines and Angles # 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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Exercise 6.1-Lines and Angles

Exercise 6.2-Lines and Angles

### 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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