Students can Download Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 3 Lines and Angles Ex 3.3

Question 1.

In Fig. 3.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer:

Arms of the ∆PQR QP and RP are produced to S and T and ∠PQT = 110°, ∠SPR = 135°. ∠PRQ = ?

Straight line PR is on straight line SQ.

∠SPR and ∠RPQ are Adjacent angles.

∴ ∠SPR + ∠RPQ = 180°

135 + ∠RPQ = 180°

∴ ∠RPQ = 180 – 135

∠RPQ = 45° (i)

Similarly QP straight line is on straight line TR. ∠TQP and ∠PQR are Adjacent angles.

∴ ∠RQP + ∠PQR = 180°

110 + ∠PQR = 180°

∠PQR = 180 – 110

∴ ∠PQR = 70°

Now, in ∆PQR,

∠QPR + ∠PQR + ∠PRQ = 180°

45 + 70 + ∠PRQ = 180°

115 + ∠PRQ = 180°

∠PRQ = 180 – 115

∴ ∠PRQ = 65°.

Question 2.

In Fig. 3.40, ∠X = 62°. ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆ XYZ, find ∠OZY and ∠YOZ

Answer:

In this figure, ∠X = 62°

∠XYZ = 54°

In ∆XYZ, YO and ZO are angular bisectors of ∠XYZ and ∠XZY.

Then, ∠OZY = ?

∠OYZ = ?

In ∆ XYZ

∠X + ∠Y + ∠Z = 180°

62 + 54 + ∠Z = 180

116 + ∠Z = 180

∠Z = 180 – 116

∴ ∠Z = 64°

YO is the angular bisector of ∠Y

∴ ∠OYZ = \(\frac{54}{2}\) = 27°

ZO is the angular bisector of ∠Z

∴ ∠OZY = \(\frac{64}{2}\) = 32°

∴ ∠OZY = 32°

Now, in ∆OYZ,

∠OYZ + ∠OZY + ∠YOZ = 180°

27 + 32 + ∠YOZ = 180

59 + ∠YOZ = 180

∠YOZ = 180 – 59

∴∠YOZ = 121°

∴ ∠OZY = 32°

∠YOZ = 121°

Question 3.

In Fig. 3.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer:

If AB || DE, ∠BAC = 35, ∠CDE = 53 then ∠DCE = ?

AB || DE, AE is the bisector.

∴ ∠BAC = ∠DEC = 35° (∵ Alternate angles)

∴ ∠DEC = 35°

Now in ∆CDE,

∠DCE + ∠CDE + ∠CED = 180°

∠DCE + 53 + 35 = 180

∠DCE + 88 = 180

∠DEC = 180 – 88

∴ ∠DCE = 92°.

Question 4.

In Fig. 3.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer:

PQ and RS straight lines intersect at T.

If ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, then ∠SQT =?

In ∆PRT,

∠RPT + ∠PRT + ∠PTR = 180°

95 + 40 + ∠PTR = 180°

135 + ∠PTR = 180

∠PTR = 180 – 135

∴ ∠PTR = 45°

∠PTR = ∠STQ = 45° (∵ Vertically opposite angles)

In ∆TSQ,

∠STQ + ∠TSQ + ∠SQT =180

45 + 75 + ∠SQT = 180

120 + ∠SQT = 180

∴ ∠SQT = 180 – 120

∴ ∠SQT = 60°.

Question 5.

In Fig. 3.43, PQ⊥PS, PQ || SR, ∠SQR = 28⊥ and ∠QRT = 65°. then find the value of x and y.

Answer:

PQ⊥PS, PQ⊥SR, ∠SQR = 28⊥, ∠QRT = 65⊥, Then x = ? y = ?

Solution: ∠QRT + ∠QRS = 180° (∵ Linear pairs)

65 + ∠QRS = 180

∠QRS = 180 – 65

∴∠QRS = 115°

In ∆SRQ,

∠QRS + ∠SQR + ∠RSQ = 180°

115 + 28 + ∠RSQ = 180

∴∠RSQ = 180 – 143

∴∠RSQ = 37

Now, ∠RSQ = ∠PQS

37° = x

∴x = 37

In ∆SPQ,

∠SPQ + ∠PSQ + ∠PQS = 180°

90 + y + 37 = 180

∴y = 180 – 127

∴y = 53°.

Question 6.

In Fig. 3.44, 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 = \(\frac{1}{2}\) ∠QPR.

Answer:

Data: Arm QR of ∆PQR is produced upto S. Angular bisectors of ∠PQR and ∠PRS meet at T.

To Prove: ∠QTR = \(\frac{1}{2}\)∠QPR

Proof: Let ∠PQR = 50°, and ∠PRS = 120°,

∠PQT = ∠TQR = 25°

∠PRT = ∠TRS = 60°

QR arm of ∆PQR is produced upto S.

∴Exterior angle ∠PRS = ∠PQR + ∠QPR

120 = 50 + ∠QPR

∴ ∠QPR =120 – 50

∠QPR = 70°

∴ ∠PRQ = 60°

Now, in ∆TRQ,

∠TQR + ∠TRQ +∠QTR = 180°

25 + 120 + ∠QTR = 180°

145 + ∠QTR = 180°

∠QTR = 180 – 145

∴ ∠QTR = 35°

Now, ∠QTR = 35° ∠QPR = 70°

∠QTR = \(\frac{70}{2}\)

∴ ∠QTR = \(\frac{10}{2}\) x ∠QPR.