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