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

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

error: Content is protected !!