**CBSE Class IX NCERT Mathematics Solutions**

*Chapter 6 - Lines and Angles*

**IXth NCERT Mathematics Textbook Exercise 6.3 Solved**

(Page 107, 108)

Q1: 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.

Ans: In ∆ PQR side QP and RQ are produced to point S and T respectively such that

__/__SPR = 135^{O},__/__PQT = 110^{O}. We have to find__/__PRQ.As

__/__SPR +__/__RPQ = 180^{O}(linear pair of angles).Or, 135

^{O}+__/__RPQ = 180^{O}Or,

__/__RPQ = 180^{O}– 135^{O}= 45^{O}Now,

__/__TPQ =__/__QPR +__/__PRQOr, 110

^{O}= 45^{O}+__/__PRQOr,

__/__PRQ = 65^{O}Q2: 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.

Ans:

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

∠X + ∠Y + ∠Z = 180º

Or, 62º + 54º + ∠Z = 180º

Or, ∠Z = 180º − 116º = 64

^{O}Now, ∠OZY = 64

^{O}/2 = 32º (as OZ is the angle bisector of ∠Z)Similarly, ∠OYZ = 54

^{O}/2 = 27^{O}Consider ΔOYZ,

∠OYZ + ∠YOZ + ∠OZY = 180º

Or, 27º + ∠YOZ + 32º = 180º

Or, ∠YOZ = 180º − 59º = 121º

Q3: In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.

Ans:

__/__DEC =

__/__BAC = 35

^{O}………. (i) [Alternate interior angles]

__/__CDE = 53

^{O}………. (ii) [given]

In ∆CDE using angle sum property we have,

__/__CDE +

__/__DEC +

__/__DCE = 180

^{O}

Or, 53

^{O}+ 35^{O }+__/__DCE = 180^{O}Or,

__/__DCE = 92^{O}Q4: In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40º, ∠RPT = 95º and ∠TSQ = 75º, find ∠SQT.

Ans: Applying angle sum property in ∆PRT we have,

__/__PTR +

__/__PRT +

__/__RPT = 180

^{O}

Or,

__/__PTR + 40^{O}+ 95^{O}= 180^{O}Or,

__/__PTR = 45^{O}Or,

__/__QTS =__/__PTR = 45^{O}[vertically opposite angles]Applying angle sum property in ∆TSQ we have,

__/__QTS +

__/__TSQ +

__/__SQT = 180

^{O}

Or,

__/__SQT + 45^{O}+ 75^{O}= 180^{O}Or,

__/__SQT = 60^{O}Q5: In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28º and ∠QRT = 65º, then find the values of

*x*and*y*.Ans:

__/__QRT =

__/__RQS +

__/__QSR [as the exterior angle is equal to the sum of the two interior opposite angles].

Or, 65

^{O}= 28^{O}+__/__QSROr,

__/__QSR = 37^{O}Given that PQ ⊥ PS i.e.

__/__QPS = 90^{O}Or, PQ || SR.

So,

__/__QPS +__/__PSR = 180^{o}[sum of the consecutive interior angles on the same side of the traversal is 180^{O}].Therefore, 90

^{O}+__/__PSR = 180^{O}Or,

__/__PSR = 90^{O}Or,

__/__PSR +__/__QSR = 90^{O}Or, y +

__/__QSR = 90^{O}Or, y + 37

^{O}= 90^{O}Or, y = 53

^{O}Now consider ∆PQS,

__/__PQS +

__/__QSP +

__/__QPS = 180

^{O}

Or, x + 53

^{O}+ 90^{O}= 180^{O}Or, x = 37

^{O}Q6: 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= ½ ∠QPR.

Ans:

__/__QTR = ½

__/__QPR

As RT and QT are the bisectors of

__/__PRS and__/__PQR respectively,Therefore,

__/__PRT =__/__TRS and__/__PQT =__/__TQR.=>

__/__TRS =__/__T +__/__TQR=> 2

__/__TRS = 2__/__T + 2__/__TQR=>

__/__PRS = 2__/__T +__/__PQR=>

__/__P +__/__PQR = 2__/__T +__/__PQROr,

__/__P = 2__/__TOr,

__/__T =__/__P/2[Exterior angle = Sum of opposite interior angle].

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