Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 6 Lines and Angles Ex 6.3 Textbook Exercise Questions and Answers.

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

Question 1.

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

Answer:

Here, ∠ SPR and ∠ PQT are exterior angles.

Then, by theorem 6.8,

∠ SPR = ∠ PQR + ∠ PRQ and

∠ PQT = ∠ QPR + ∠ PRQ

∴ ∠ PQR + ∠ PRQ = 135° and

∠ QPR + ∠ PRQ = 110°

Adding these two equations,

∠ PQR + ∠ PRQ + ∠ QPR + ∠ PRQ = 135° + 110°

∴ 180° + ∠ PRQ = 245° (Theorem 6.7)

∴ ∠ PRQ = 245° – 180°

∴ ∠ PRQ = 65°

Question 2.

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

Answer:

In ∆ XYZ,

∠ X + ∠ XYZ + ∠ XZY = 180° (Theorem 6.7)

∴ 62° + 54° + ∠ XZY = 180°

∴ ∠ XZY = 180° – 62° – 54°

∴ ∠ XZY = 64°

YO and ZO are bisectors of ∠ XYZ and ∠ XZY respectively.

∴ ∠ OYZ = \(\frac{1}{2}\) ∠ XYZ = \(\frac{1}{2}\) × 54° = 27° and

∠ OZY = \(\frac{1}{2}\) ∠ XZY = \(\frac{1}{2}\) × 64° = 32°.

Now, in ∆ OYZ,

∠ OYZ + ∠ OZY + ∠ YOZ = 180° (Theorem 6.7)

∴ 27° + 32° + ∠ YOZ = 180°

∴ ∠ YOZ = 180° – 27° – 32°

∴ ∠ YOZ = 121°

Question 3.

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

Answer:

AB || DE and AE is transversal for them.

∴ ∠ AED = ∠ BAE (Alternate interior angles)

∴ ∠ CED = ∠ BAC (Point C lies on line AE)

∴ ∠ CED = 35° (Given : ∠ BAC = 35°)

In ∆ CDE, by theorem 6.8

∠ CDE + ∠ CED + ∠ DCE = 180°

∴ 53° + 35° + ∠ DCE = 180°

∴ ∠ DCE = 180° – 53° – 35°

∴ ∠ DCE = 92°

Question 4.

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

Answer:

In ∆ PRT,

∠ RPT + ∠ PRT + ∠ PTR = 180° (Theorem 6.7)

∴ 95° + 40° + ∠ PTR = 180°

∴ 135° + ∠ PTR = 180°

∴ ∠ PTR = 180°- 135°

∴ ∠ PTR = 45°

Lines PQ and RS intersect at point T.

∴ ∠ STQ = ∠ PTR (Vertically opposite angles)

∴ ∠ STQ = 45°

In ∆ STQ,

∠ TSQ + ∠ STQ + ∠ SQT = 180° (Theorem 6.7)

∴ 75° + 45° + ∠ SQT = 180°

∴ 120° + ∠ SQT = 180°

∴ ∠ SQT = 60°

Question 5.

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

Answer:

PQ || SR and QR is transversal for them.

∴ ∠ PQR = ∠ QRT (Alternate interior angles)

∴ ∠ PQR = 65° (Given : ∠ QRT = 65°)

∴ ∠ PQS + ∠ SQR = 65° (Adjacent angles)

∴ x + 28° = 65° (Given : ∠ SQR = 28°)

∴ x = 65° – 28°

∴ x = 37°

PQ ⊥ PS

∴ ∠ SPQ = 90°

In ∆ PSQ,

∠ SPQ + ∠ PQS + ∠ PSQ = 180° (Theorem 6.7)

∴ 90° + 37° + y = 180°

∴ 127° + y = 180°

∴ y = 180°- 127°

∴ y = 53°

Question 6.

In the given 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.

Answer:

QT is the bisector of ∠ PQR and RT is the bisector of ∠ PRS.

∴ ∠ TQR = \(\frac{1}{2}\) ∠ PQR and ∠ TRS = \(\frac{1}{2}\) ∠ PRS ……………… (1)

∠ PRS is an exterior angle of ∆ PQR.

∴ ∠ PRS = ∠ QPR + ∠ PQR

∴ \(\frac{1}{2}\) ∠ PRS = \(\frac{1}{2}\) ∠ QPR + \(\frac{1}{2}\) ∠ PQR

∴ ∠ TRS = \(\frac{1}{2}\) ∠ QPR + ∠ TQR [By (1)] …………… (2)

In ∆ TQR, ∠ TRS is an exterior angle.

∴ ∠ TRS = ∠ QTR + ∠ TQR ……………. (3)

From (2) and (3), we get

∠ QTR + ∠ TQR = \(\frac{1}{2}\) ∠ QPR + ∠ TQR

∴ ∠ QTR = \(\frac{1}{2}\) ∠ QPR