## MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.

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

Solution:

Given

∠SPR = 135°

∠PQT = 110°

To find. ∠PRQ

Calculation:

∠Q = 110°

110° + ∠PQR = 180° (LPA’s)

∠1 = 70°

∠P = ∠1 + ∠PRQ

135° = 70° + ∠PRQ

135° – 70° = ∠PRQ

65° = ∠PRQ

Question 2.

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

Solution:

Given

∠X = 62°, ∠XYZ = 54°

∠OYZ = \(\frac{1}{2}\)∠XYZ

∠OZY = \(\frac{1}{2}\)∠XZY

To find ∠OZY and ∠YOZ

Calculation:

In ∠XYZ

∠X + ∠Y + ∠Z = 180° (ASP)

62° + 54° + ∠Z = 180°

116° + ∠Z = 180°

∴ ∠Z = 64°

In ∠OYZ

∠OYZ + ∠OZY + ∠O = 180° (ASP)

⇒ \(\frac{1}{2}\)∠XYZ + \(\frac{1}{2}\)∠XZY + ∠O = 180°

32° + 27° + ∠O = 180°

59° + ∠O = 180°

∠O = 121°

Question 3.

In Fig. given below, if AS ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:

Given

AS ∥ DE,

∠BAC = 35°

∠CDE = 53°

To find. ∠DCE

Calculation:

AB ∥ DE and AE is the transversal.

∠BAE = ∠AED = 35°

In ∆DEC

∠D + ∠E + ∠C = 180°

53° + 35° + ∠C = 180°

88° + ∠C = 180°

∠C = 180° – 88° = 92°

Question 4.

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

Solution:

Given

∠PRT = 40°

∠RPT = 95°

∠TSQ = 75°

To find. ∠SQT

Calculation:

In ∆PRT

∠P + ∠R + ∠T= 180° (ASP)

95° + 40 ° + ∠T= 180°

∠T = 180° – 135° = 45°

∠T = ∠STQ = 45°

In ∆TSQ

∠STQ + ∠S + ∠Q = 180° (V.O.A’S)

45° + 75° + ∠Q = 180° (ASP)

∠Q = 180° – 120° = 60°

Question 5.

In Fig. given below, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:

Given

PQ ⊥ PS

PQ ∥ SR

∠SQR = 28°

∠QRT = 65°

To find, x and y

Calculation:

PQ ∥ SR and QR is the transversal

x + 28° = 65° (A.I.A’s)

x = 65° – 28° = 37°

In ∆PQS

x + y + 90° = 180° (ASP)

37° + y + 90° = 180°

∴ y = 180° – 127° = 53°

Question 6.

In Fig. given below, the side QR to ∆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.

Solution:

Given

∠TQR = \(\frac{1}{2}\)∠PQR

∠TRS = \(\frac{1}{2}\)∠PRS

To prove. ∠QTR = \(\frac{1}{2}\)∠QPR

Proof:

In ∆PQR

∠PRS = ∠P + ∠Q (EAP)

In ∆TQR, ∠TRS = ∠T + ∠TQR (EAP)

\(\frac{1}{2}\)∠PRS = ∠T + \(\frac{1}{2}\)∠PQR

Multiplying both sides by 2

∠PRS = 2∠T+ ∠PQR

From (1) and (2) we get,

∠P + ∠Q = 2∠T + ∠PQR

∠P=2∠T

∠T= \(\frac{1}{2}\)∠P

∠QTR = \(\frac{1}{2}\)∠QPR

Proved.