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

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

Question 1.

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

Solution:

∠PQT + ∠PQR = 180° (Linear pair)

⇒ 110° + ∠PQR = 180°

⇒ ∠PQR = 180° – 110°

⇒ ∠PQR = 70°

∠RPS = ∠PQR + ∠PRQ (Exterior angle property)

⇒ 135° – 70° + ∠PRQ

⇒ ∠PRQ = 135° – 70°

∠PRQ = 65°

Question 2.

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

Solution:

In ΔXYZ,

∠YXZ + ∠XYZ + ∠XZY = 180° (Angle sum property)

⇒ 62° + 54° + ∠XZY = 180°

⇒ 116° + ∠XZY = 180°

⇒ ∠XZY = 180° – 116°

∠XZY = 64°

∠OZY = \(\frac {∠XZY}{2}\) = \(\frac {64°}{2}\) (OZ is bisector of ∠XZY)

∠OYZ = 32°

∠OZY = \(\frac {∠XZY}{2}\)

(OY is bisector of ∠XYZ)

∠OYZ = \(\frac {54°}{2}\)

∠OYZ = 27°

In ΔOYZ,

∠OYZ + ∠OZY + ∠YOZ = 180°

(Angle sum property)

⇒ 27° + 32° + ∠YOZ = 180°

⇒ 59° + ∠YOZ = 180°

⇒ ∠YOZ = 180° – 59°

⇒ ∠YOZ = 121°

Question 3.

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

Solution:

AB || DE

∠DEC = ∠BAC

(Alternate interior angles)

⇒ ∠DEC = 35°

Now in ACDE,

∠DCE + ∠CDE + ∠DEC = 180° (Angle sum property)

∠DCE + 53° + 35° = 180°

∠DCE = 180° – 88° ∠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 ΔPRT,

∠PRT + ∠RPT + ∠PTR = 180° (Angle sum property)

40° + 95° + ∠PTR = 180°

135° + ∠PTR = 180°

PTR = 180° – 135° ∠PTR = 45°

∠STQ = ∠PTR

(Vertically opposite angles)

∠STQ = 45°

Now in ΔSQT,

∠STQ + ∠SQT ∠QST = 180° (Angle sum property)

⇒ 45° + ∠SQT + 75° = 180°

⇒ ∠SQT + 120° – 180°

⇒ ∠SQT = 180° – 120°

⇒ ∠SQT = 60°

Question 5.

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

Solution:

We have PQ || SR

∴ PQ || ST

∴ ∠PQR = ∠QTR (Alternate interior angles)

⇒ x + 28° = 65°

⇒ x = 65° – 28°

⇒ x = 37°

In ΔPQS,

∠PQS + ∠PSQ + ∠SPQ = 180° (Angle sum property)

⇒ x + y + 90° = 180°

37° + 90° + y = 180°

127° + y = 180°

⇒ y= 180° – 127°

⇒ y = 53°

Question 6.

In 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 = \(\frac {1}{2}\) ∠QPR.

Solution:

Given: ΔPQR in which side QR produced to S and bisectors of ∠QPRPQR and ∠QPRPRS meet at T.

To Prove: ∠QPR = \(\frac {1}{2}\) ∠QPR

Proof: ∠TRS is the exterior angle of ΔTQR.

∴ ∠TRS = ∠1 + ∠2 …….(1)

(Exterior angle of triangle is equal to sum of two interior opposite angles)

∠PRS is an exterior angle of ΔPQR.

∴ ∠PRS = ∠PQR + ∠QPR

(Exterior angle of a triangle is equal to the sum of its two interior opposite angles)

⇒ 2 ∠TRS = 2∠1 + ∠QPR

2(∠1 + ∠2) = 2∠1 + ∠QPR [From eqn(Δ)]

⇒ 2∠1 + 2∠2 = 2∠1 + ∠QPR

⇒ 2∠2 = ∠QPR

⇒ ∠2 = \(\frac {1}{2}\) ∠QPR

⇒ ∠QTR = \(\frac {1}{2}\) ∠QPR