EXERCISE 6.3
Q 1, Ex. 6.3, Page No 107, Lines & Angles, Class 9th Maths
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
Q 2, Ex. 6.3, Page No 107, Lines & Angles - Class 9th Maths
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 = $\frac{64}{2}$ = 32º (OZ is the angle bisector of ∠XZY)
Similarly, ∠OYZ = $\frac{54}{2}$= 27º
Using angle sum property for ΔOYZ, we obtain
∠OYZ + ∠YOZ + ∠OZY = 180º
27º + ∠YOZ + 32º = 180º
∠YOZ = 180º − 59º
∠YOZ = 121º
Q 3, Ex. 6.3, Page No 107, Lines & Angles - Class 9th Solutions
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
Q 4, Ex. 6.3, Page No 107, Lines & Angles, Class 9th Maths
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
Q 5, Ex. 6.3, Page No 108, Lines & Angles, Class 9th Maths
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
x = 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
y = 53º
x = 37º and y = 53º
Q 6, Ex. 6.3, Page No 108, Lines & Angles, NCERT Class 9th Maths
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=$\frac{1}{2}$∠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 = $\frac{1}{2}$∠QPR
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