**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 Î”PQR are produced to points S and T respectively. If ∠SPR = 135Âº and ∠PQT = 110Âº, 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 + __/ __PRQ

Or, 110^{O} = 45^{O} + __/ __PRQ

Or, __/ __PRQ = 65^{O}

Q2: In the given figure, ∠X = 62Âº, ∠XYZ = 54Âº. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Î”XYZ, find ∠OZY and ∠YOZ.

Ans:

As the sum of all interior angles of a triangle is 180Âº, therefore, for Î”XYZ,

∠X + ∠Y + ∠Z = 180Âº

Or, 62Âº + 54Âº + ∠Z = 180Âº

Or, ∠Z = 180Âº − 116Âº = 64^{O}

Now, ∠OZY = 64^{O}/2 = 32Âº (as OZ is the angle bisector of ∠Z)

Similarly, ∠OYZ = 54^{O}/2 = 27^{O}

Consider Î”OYZ,

∠OYZ + ∠YOZ + ∠OZY = 180Âº

Or, 27Âº + ∠YOZ + 32Âº = 180Âº

Or, ∠YOZ = 180Âº − 59Âº = 121Âº

Q3: In the given figure, if AB || DE, ∠BAC = 35Âº and ∠CDE = 53Âº, 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 = 40Âº, ∠RPT = 95Âº and ∠TSQ = 75Âº, 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 = 28Âº and ∠QRT = 65Âº, 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} + __/ __QSR

Or, __/ __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 Î”PQR 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 + __/ __PQR

Or, __/ __P = 2__/ __T

Or, __/ __T = __/ __P/2

[Exterior angle = Sum of opposite interior angle].

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