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NCERT solution class 9 chapter 6 Lines and Angles exercise 6.3 mathematics

EXERCISE 6.3


Page No 107:

Question 1:

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

Answer:

It is given that,

∠SPR = 135º and ∠PQT = 110º

∠SPR + ∠QPR = 180º (Linear pair angles)

⇒ 135º + ∠QPR = 180º

⇒ ∠QPR = 45º

Also, ∠PQT + ∠PQR = 180º (Linear pair angles)

⇒ 110º + ∠PQR = 180º

⇒ ∠PQR = 70º

As the sum of all interior angles of a triangle is 180º, therefore, for ΔPQR,

∠QPR + ∠PQR + ∠PRQ = 180º

⇒ 45º + 70º + ∠PRQ = 180º

⇒ ∠PRQ = 180º − 115º

⇒ ∠PRQ = 65º


Question 2:

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

Answer:

As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

∠X + ∠XYZ + ∠XZY = 180º

62º + 54º + ∠XZY = 180º

∠XZY = 180º − 116º

∠XZY = 64º

∠OZY =  = 32º (OZ is the angle bisector of ∠XZY)

Similarly, ∠OYZ = = 27º

Using angle sum property for ΔOYZ, we obtain

∠OYZ + ∠YOZ + ∠OZY = 180º

27º + ∠YOZ + 32º = 180º

∠YOZ = 180º − 59º

∠YOZ = 121º


Question 3:

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

Answer:

AB || DE and AE is a transversal.

∠BAC = ∠CED (Alternate interior angles)

∴ ∠CED = 35º

In ΔCDE,

∠CDE + ∠CED + ∠DCE = 180º (Angle sum property of a triangle)

53º + 35º + ∠DCE = 180º

∠DCE = 180º − 88º

∠DCE = 92º


Question 4:

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

Answer:

Using angle sum property for ΔPRT, we obtain

∠PRT + ∠RPT + ∠PTR = 180º

40º + 95º + ∠PTR = 180º

∠PTR = 180º − 135º

∠PTR = 45º

∠STQ = ∠PTR = 45º (Vertically opposite angles)

∠STQ = 45º

By using angle sum property for ΔSTQ, we obtain

∠STQ + ∠SQT + ∠QST = 180º

45º + ∠SQT + 75º = 180º

∠SQT = 180º − 120º

∠SQT = 60º


Page No 108:

Question 5:

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

Answer:

It is given that PQ || SR and QR is a transversal line.

∠PQR = ∠QRT (Alternate interior angles)

x + 28º = 65º

= 65º − 28º

x = 37º

By using the angle sum property for ΔSPQ, we obtain

∠SPQ + x + y = 180º

90º + 37º + y = 180º

y = 180º − 127º

= 53º

x = 37º and y = 53º


Question 6:

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

Answer:

In ΔQTR, ∠TRS is an exterior angle.

∠QTR + ∠TQR = ∠TRS

∠QTR = ∠TRS − ∠TQR (1)

For ΔPQR, ∠PRS is an external angle.

∠QPR + ∠PQR = ∠PRS

∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)

∠QPR = 2(∠TRS − ∠TQR)

∠QPR = 2∠QTR [By using equation (1)]

∠QTR = ∠QPR

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