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

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.
In the figure, sides QP and RQ of APQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110% find ∠PRQ.

Answer:
We have :
∠QPR + ∠SPR = 180° (Linear pair)
∠QPR + 135° = 180°
or ∠QPR = 180° – 135° = 45°
Now, ∠TQP = ∠QPR + ∠PRQ [By exterior angle theorem (8)]
⇒ 110° = 45° + ∠PRQ
⇒ ∠PRQ a 110° – 45° = 65°
Hence, ∠PRQ = 65°.

Question 2.
In the figure, ∠X = 62% ∠XYZ = 54°. If YO and ZO are ‘ the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.

Answer:
Consider ∆XYZ.
∠YXZ + ∠XYZ + ∠XZY = 180° (Angle-sum property)
⇒ 62° + 54° + ∠XZY = 180° (∵ ∠YXZ = 62% ∠XYZ = 54°)
⇒ ∠XZY = 180° – 62° – 54° = 64°
Since YO and ZO are bisectors of ∠XYZ and ∠XZY, therefore
∠OYZ = \(\frac{1}{2}\) × ∠XYZ = \(\frac{1}{2}\) × 54° = 27°
and, ∠OZY = \(\frac{1}{2}\) × ∠XZY = \(\frac{1}{2}\) × 64° = 32°

Now, in ∆OYZ, we have :
∠YOZ + ∠OYZ + ∠OZY = 180° (Angle -sum property)
∴ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° – 27° – 32° = 121°
Hence, ∠OZY = 32° and ∠YOZ = 121°.

Question 3.
In the figure, if AB ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer:
Since AB ∥ DE and transversal AB intersects them at A and E respectively, therefore
∠DEA = ∠BAE (Alternate 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 the figure, if lines PQ and RS intersect at point r, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer:
In ∆PRT, we have
∠PRT + ∠RTP + ∠TPR = 180°
(Angle-sum property)
⇒ 40° + ∠RTP + 95° = 180°
⇒ ∠RTP = 180° – 40° – 95° = 45°
Now, ∠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 the figure, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Answer:
Using exterior angle property in ASRQ, we have :
∠QRT = ∠RQS + ∠QSR
⇒ 65° = 28° + ∠QSR (∵ ∠QRT = 65°, ∠RQS = 28°)
⇒ ∠QSR = 65° – 28° = 37°
Now, PQ ∥ SR and the transversal PS intersects them at P and S respectively.
∠PSR + ∠SPQ = 180° (Sum of the interior angles on the same side of the transversal is 180°.)
⇒ (∠PSQ + ∠QSR) + 90° = 180°
⇒ y + 37° + 90° = 180°
⇒ y = 180° – 90° – 37° = 53°

In the right triangle SPQ, we have :
∠PQS + ∠PSQ = 90°
⇒ x + 53° = 90°
⇒ x = 90° – 53° = 37°
Hence, x = 37° and y = 53°.

Question 6.
In the figure, the side QR of APQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ~ ∠QPR.

Answer:
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 = ∠TQR + ∠T
⇒ \(\frac{1}{2}\)∠P = ∠T
⇒ ∠QTR = \(\frac{1}{2}\)∠QPR.
Hence Proved.