These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3

Question 1.

In Fig. 6.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135 and ∠PQT = 110°, Find ∠PRQ.

Solution:

In ∆PQR

∠SPR + ∠RPQ = 180° (Linear pair)

or, ∠RPQ = 180° – 135° (∵ ∠SPR = 135°)

or, ∠RPQ = 45° …….(i)

Again, ∠PQT + ∠PQR = 180° (Linear pair)

or, 110° + ∠PQR = 180° (∵ ∠PQT = 110°)

∠PQR = 180° – 110°

or, ∠PQR = 70° ……(ii)

Now, we know that sum of all angles of a ∆ is 180°

Therefore,

∠RPQ + ∠PQR + ∠PRQ = 180°

or, 45° + 70° + ∠RPQ = 180° [From equ. (i) and (ii)]

∠RPQ = 180° – 45° – 70° = 65°

Therefore, ∠RPQ = 65°

Question 2.

In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisector of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.

Solution:

In ∆XYZ,

∠X + ∠XYZ + ∠XZY = 180° (Angle sum property of ∆)

or, 62° + 54° + XZY = 180° (∴ ∠X = 65° & ∠XYZ = 54°)

or, ∠XYZ = 180° – 116°

or, ∠XZY = 64° ….(i)

Again, ∠XZY = ∠XZO + ∠OZY (∴ OZ is the bisector of ∠XZY)

or, 2∠OZY = 64°

or, ∠OZY = 32°

∠OYZ + ∠OZY + ∠YOZ = 180° (Angle sum property of ∆)

27° + 32° + ∠YOZ = 180°

or, ∠YOZ = 121°

Question 3.

In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:

We have given AB || DE

∠BAC = ∠DEC (Pair of alternate interior angles)

or, ∠DEC = 35° (∵ ∠BAC = 35°)

Now, In ∆CDE

∠CDE + ∠CED + ∠DCE = 180° (Angle sum property of ∆)

⇒ 53 + 35 + ∠DCE = 180°

⇒ ∠DCE = 180° – 53° – 35°

⇒ ∠DCE = 92°

Question 4.

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

Solution:

In ∆PRT,

∠RPT + ∠PRT + ∠PTR = 180° (Angle sum property of ∆)

95 + 40 + ∠PTR = 180°

or, ∠PTR = 180° – 95° – 40°

or, ∠PTR = 45°

Again, ∠PTR = ∠QTS (Vertically opposite angles)

∴ ∠QTS = 45°

Now, In ∆QTS,

∠QTS + ∠TSQ + ∠SQT = 180° (Angle sum property of ∆)

45° + 75° + ∠SQT = 180° (∵ ∠QTS = 45° and ∠TSQ = 75°)

or, ∠SQT = 180° – 45° – 75°

∴ ∠SQT = 60°

Question 5.

In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the value of x and y.

Solution:

We have given PQ || SR

∴ ∠PQR = ∠QRT (Pair of alternate interior angles)

⇒ x + 28° = 65°

⇒ x = 65° – 28°

⇒ x = 37°

Now, in ∆PQS

∠QPS + x + y = 180° (Angle sum property of ∆)

⇒ 90° + 37° + y = 180° (∴ PQ ⊥ PS and x = 37)

⇒ y = 180° – 90° – 37°

⇒ y = 53°

Question 6.

In Fig. 6.44, 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.

Solution:

In ∆PQR, side QR is produced to S, so by exterior angle property,

∠PRS = ∠P + ∠PQR

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

⇒ ∠TRS = \(\frac {1}{2}\) ∠P + ∠TQR ……(1)

[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]

Now, in ∆QRT, we have

∠TRS = ∠TQR + ∠T ……(2)

[Exterior angle property of a triangle]

From (1) and (2),

we have ∠TQR + \(\frac {1}{2}\) ∠P = ∠TQR + ∠T

⇒ \(\frac {1}{2}\) ∠P = ∠T

⇒ \(\frac {1}{2}\) ∠QPR = ∠QTR or ∠QTR = \(\frac {1}{2}\) ∠QPR