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 Î”PQR are produced to points S and T respectively. If âˆ SPR = 135Â° and âˆ PQT = 110Â°, find âˆ PRQ.

Solution:

âˆ PQT + âˆ PQR = 180Â° (Linear pair)

â‡’ 110Â° + âˆ PQR = 180Â°

â‡’ âˆ PQR = 180Â° – 110Â°

â‡’ âˆ PQR = 70Â°

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

â‡’ 135Â° – 70Â° + âˆ PRQ

â‡’ âˆ PRQ = 135Â° – 70Â°

âˆ PRQ = 65Â°

Question 2.

In figure âˆ X = 62Â°, âˆ XYZ = 54Â°. If YO and ZO are the bisectors of âˆ XYZ and âˆ XZY respectively of Î”XYZ, find âˆ OZY and âˆ YOZ.

Solution:

In Î”XYZ,

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

â‡’ 62Â° + 54Â° + âˆ XZY = 180Â°

â‡’ 116Â° + âˆ XZY = 180Â°

â‡’ âˆ XZY = 180Â° – 116Â°

âˆ XZY = 64Â°

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

âˆ OYZ = 32Â°

âˆ OZY = \(\frac {âˆ XZY}{2}\)

(OY is bisector of âˆ XYZ)

âˆ OYZ = \(\frac {54Â°}{2}\)

âˆ OYZ = 27Â°

In Î”OYZ,

âˆ OYZ + âˆ OZY + âˆ YOZ = 180Â°

(Angle sum property)

â‡’ 27Â° + 32Â° + âˆ YOZ = 180Â°

â‡’ 59Â° + âˆ YOZ = 180Â°

â‡’ âˆ YOZ = 180Â° – 59Â°

â‡’ âˆ YOZ = 121Â°

Question 3.

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

Solution:

AB || DE

âˆ DEC = âˆ BAC

(Alternate interior angles)

â‡’ âˆ DEC = 35Â°

Now in ACDE,

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

âˆ DCE + 53Â° + 35Â° = 180Â°

âˆ DCE = 180Â° – 88Â° âˆ DCE = 92Â°

Question 4.

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

Solution:

In Î”PRT,

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

40Â° + 95Â° + âˆ PTR = 180Â°

135Â° + âˆ PTR = 180Â°

PTR = 180Â° – 135Â° âˆ PTR = 45Â°

âˆ STQ = âˆ PTR

(Vertically opposite angles)

âˆ STQ = 45Â°

Now in Î”SQT,

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

â‡’ 45Â° + âˆ SQT + 75Â° = 180Â°

â‡’ âˆ SQT + 120Â° – 180Â°

â‡’ âˆ SQT = 180Â° – 120Â°

â‡’ âˆ SQT = 60Â°

Question 5.

In figure, if PQ âŠ¥PS, PQ || SR, âˆ SQR = 28Â° and âˆ QRT = 65Â°, then find the values of x and y.

Solution:

We have PQ || SR

âˆ´ PQ || ST

âˆ´ âˆ PQR = âˆ QTR (Alternate interior angles)

â‡’ x + 28Â° = 65Â°

â‡’ x = 65Â° – 28Â°

â‡’ x = 37Â°

In Î”PQS,

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

â‡’ x + y + 90Â° = 180Â°

37Â° + 90Â° + y = 180Â°

127Â° + y = 180Â°

â‡’ y= 180Â° – 127Â°

â‡’ y = 53Â°

Question 6.

In figure, the side QR of APQR is produced to a point S. If the bisectors of âˆ PQR and âˆ PRS meet at point T, then prove that âˆ QTR = \(\frac {1}{2}\) âˆ QPR.

Solution:

Given: Î”PQR in which side QR produced to S and bisectors of âˆ QPRPQR and âˆ QPRPRS meet at T.

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

Proof: âˆ TRS is the exterior angle of Î”TQR.

âˆ´ âˆ TRS = âˆ 1 + âˆ 2 …….(1)

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

âˆ PRS is an exterior angle of Î”PQR.

âˆ´ âˆ PRS = âˆ PQR + âˆ QPR

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

â‡’ 2 âˆ TRS = 2âˆ 1 + âˆ QPR

2(âˆ 1 + âˆ 2) = 2âˆ 1 + âˆ QPR [From eqn(Î”)]

â‡’ 2âˆ 1 + 2âˆ 2 = 2âˆ 1 + âˆ QPR

â‡’ 2âˆ 2 = âˆ QPR

â‡’ âˆ 2 = \(\frac {1}{2}\) âˆ QPR

â‡’ âˆ QTR = \(\frac {1}{2}\) âˆ QPR