Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3 Textbook Questions and Answers.

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

Question 1.

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

Solution:

We have,

∠QPR + ∠SPR = 180°

⇒ ∠QPR + 135° = 180°

∠QPR = 180° – 135° = 45°

Now, ∠TQP = ∠QPR + ∠PRQ [By exterior angle theorem]

⇒ 110° = 45° + ∠PRQ

⇒ ∠PRQ = 110° – 45° = 65°

Hence, ∠PRQ = 65°

Question 2.

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

Solution:

Consider ∆ XYZ,

∠YXZ + ∠XYZ + ∠XZY = 180° [Angle-sum property]

⇒ 62° + 54° + ∠X∠y

[∵∠YXZ = 62°, ∠XYZ = 54°]

⇒ ∠ XZY = 180°- 62° – 54° = 64°

Since YO and ZO are Therefore ∠XYZ and ∠XZY.

Therefore

Question 3.

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

Solution:

Since AB || DE and transversal AE intersect them at A and E respectively.

∴ ∠DEA = ∠BAE [Alternative angles]

⇒ ∠DEC = 35° [∵ ∠DEA = ∠DEC and ∠BAE = 35°]

In ∆ DEC, we have

∠DCE + ∠DEC + ∠CDE = 180° [Angle-sum property]

⇒ ∠DCE + 35° + 53° = 180°

⇒ ∠DCE = 180° – 35° – 53° = 92°

Hence, ∠DCE = 92°.

Question 4.

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

Solution:

In ∆ PRT, we have

∠PRT + ∠RTP + ∠TPR = 180° [Angle-sum Property]

⇒ 40° + ∠RTP + 95° = 180°

⇒ ∠RTP = 180° – 40° – 95° = 45°

⇒ ∠STQ = ∠RTP [Vertically opp. angles]

⇒ ∠STQ = 45° [∵∠RTP = 45°(proved) ]

In ∆ TQS, we have

∠SQT + ∠STQ + ∠TSQ = 180° [Angle-sum Property]

⇒ ∠SQT + 45° + 75° = 180° [ ∵ ∠STQ = 45° (proved) ]

⇒ ∠SQT = 180° – 45° – 75° = 60°

Hence, ∠SQT = 60°.

Question 5.

In figure, if PQ ⊥ PS, P PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:

Using exterior angle property in ∆ SRQ, we have

∠QRT = ∠RQS + ∠QSR

⇒ 65° = 28° + ∠QSR

[∵ ∠QRT = 65°, ∠RQS = 28°]

⇒ QSR = 65° – 28°

= 37°

Since PQ || SR and the transversal PS intersects them at P and S respectively.

∴ ∠PSR + ∠SPQ = 180° [Sum of consecutive interior angles is 180°]

⇒ (∠PSQ + ∠QSR) + 90° = 180°

⇒ y + 37° + 90° = 180°

⇒ y = 180° – 90° – 37°

= 53°

In the right ∆ SPQ, we have

∠PQS + ∠PSQ = 90°

⇒ x + 53° = 90°

⇒ x = 90° – 53°

= 37°

Hence, x = 37°

and y = 53°

Question 6.

In figure, 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, we have

ext. ∠PRS = ∠P + ∠Q

⇒ \(\frac { 1 }{ 2 }\) ext. ∠PRS = \(\frac { 1 }{ 2 }\)∠P + \(\frac { 1 }{ 2 }\)∠Q

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

[∵ QT and RT are bisectors of ∠Q and ∠PRS respectively ∴ ∠Q = 2∠TQR and ext. ∠PRS = 2∠TRS]

In ∆ QRT, we have

ext. ∠TRS = ∠TQR + ∠T … (2)

From (1) and (2), we get

\(\frac { 1 }{ 2 }\)∠P + ∠TQR + ∠T

⇒ \(\frac { 1 }{ 2 }\)∠P = ∠T

⇒ ∠QTR = \(\frac { 1 }{ 2 }\)∠QPR