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

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