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# Class 9 NCERT Solutions – Chapter 6 Lines And Angles – Exercise 6.3

• Last Updated : 11 Dec, 2020

### Question1: 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:

Given: âˆ TQP = 110Â°, âˆ SPR = 135Â°

TQR is a Straight line as we can see in the figure

As we have studied in this chapter, TQP and PQR will form a linear pair

â‡’ âˆ TQP + âˆ PQR = 180Â°  ———-(i)

Putting the value of âˆ TQP = 110Â° in Equation (i) we get,

â‡’ 110Â° + âˆ PQR = 180Â°

â‡’ âˆ PQR = 70Â°

Consider the Î”PQR,

Here, the side QP is extended to S and so, SPR forms the exterior angle.

Thus, âˆ SPR (âˆ SPR = 135Â°) is equal to the sum of interior opposite angles. (Triangle property)

Or, âˆ PQR + âˆ PRQ = 135Â° ———(ii)

Now, putting the value of PQR = 70Â° in equation (ii) we get,

âˆ PRQ = 135Â° – 70Â°

Hence, âˆ PRQ = 65Â°

### Question 2: In Figure, âˆ X = 62Â°, âˆ  XYZ = 54Â°. If YO and ZO are the bisectors of XYZ and XZY respectively of Î” XYZ, find OZY and YOZ.

Solution:

Given: âˆ X = 62Â°, âˆ XYZ = 54Â°

As we have studied in this chapter,

We know that the sum of the interior angles of the triangle is 180Â°.

So, âˆ X +âˆ XYZ +âˆ XZY = 180Â°

Putting the values as given in the question we get,

62Â°+54Â° + âˆ XZY = 180Â°

Or, âˆ XZY = 64Â°

Now, we know that ZO is the bisector so,

âˆ OZY = Â½ XZY

âˆ´ âˆ OZY = 32Â°

Similarly, YO is a bisector and so,

âˆ OYZ = Â½ XYZ

Or, âˆ OYZ = 27Â° (As XYZ = 54Â°)

Now, as the sum of the interior angles of the triangle,

âˆ OZY +âˆ OYZ +O = 180Â°

Putting their respective values, we get,

âˆ O = 180Â°-32Â°-27Â°

Hence, âˆ O = 121Â°

### Question 3: In Figure, if AB || DE, âˆ BAC = 35Â° and âˆ CDE = 53Â°, find âˆ DCE.

Solution:

Given: AB || DE, âˆ BAC = 35Â° and âˆ CDE = 53Â°

Since, we know that AE is a transversal of AB and DE

Here, BAC and AED are alternate interior angles.

Hence, âˆ BAC = âˆ AED

âˆ BAC = 35Â° (Given)

âˆ AED = 35Â°

Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180Â°.

âˆ´ âˆ DCE + âˆ CED + âˆ CDE = 180Â°

Putting the values, we get

âˆ DCE + 35Â° + 53Â° = 180Â°

Hence, 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:

Given: âˆ PRT = 40Â°, âˆ RPT = 95Â° and âˆ TSQ = 75Â°

In â–³PRT.

âˆ PRT +âˆ RPT + âˆ PTR = 180Â° (The Sum of all the angles of Triangle is 180Â°)

â‡’ âˆ PTR = 45Â°

Now âˆ PTR will be equal to STQ as they are vertically opposite angles.

â‡’ âˆ PTR = âˆ STQ = 45Â°

Again, in triangle STQ,

â‡’ âˆ TSQ +âˆ PTR + âˆ SQT = 180Â°  (The Sum of all the angles of Triangle is 180Â°)

Solving this we get,

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:

Given: PQ âŠ¥ PS, PQ || SR, âˆ SQR = 28Â° and âˆ QRT = 65Â°

x + SQR = QRT (As they are alternate angles since QR is transversal)

Now, Putting the value of âˆ SQR = 28Â° and âˆ QRT = 65Â°

â‡’ x + 28Â° = 65Â°

âˆ´ x = 37Â°

It is also known that alternate interior angles are same and

â‡’ QSR = x = 37Â°

Also,

â‡’ QRS + QRT = 180Â° (As they form a Linear pair)

Putting the value of âˆ QRT = 65Â° we get,

â‡’ QRS + 65Â° = 180Â°

â‡’ QRS = 115Â°

As we know that the sum of the angles in a quadrilateral is 360Â°.

â‡’ P + Q + R + S = 360Â°

Putting their respective values, we get,

â‡’ S = 360Â° – 90Â° – 65Â° – 115Â° = 900

In Î” SPQ

â‡’ âˆ SPQ + x + y = 1800

â‡’ 900 + 370 + y = 1800

â‡’ y = 1800 â€“ 1270 = 530

Hence, y = 53Â°

### Question 6: In Figure, the side QR of Î”PQR is produced to a point S. If the bisectors of âˆ PQR and âˆ PRS meet at point T, then prove that âˆ QTR = Â½ âˆ QPR.

Solution:

Given: T is the bisector of âˆ PQR and âˆ PRS,

To Prove: âˆ QTR = Â½ âˆ QPR

Proof:

Consider the Î”PQR.

âˆ PRS is an exterior angle.

âˆ QPR and âˆ PQR are interior angles.

â‡’ âˆ PRS = âˆ QPR + âˆ PQR (According to triangle property)

â‡’ âˆ PRS – âˆ PQR = âˆ QPR  ————(i)

Now, consider the Î”QRT,

âˆ TRS = âˆ TQR + âˆ QTR (Since exterior angle are equal)

â‡’ âˆ QTR = âˆ TRS – âˆ TQR

We know that QT and RT bisect âˆ PQR and âˆ PRS respectively.

So, âˆ PRS = 2 âˆ TRS and âˆ PQR = 2âˆ TQR

Now,

â‡’ âˆ QTR = Â½ âˆ PRS â€“ Â½ âˆ PQR

â‡’ âˆ QTR = Â½ âˆ (PRS – PQR)

From equation (i) we know that âˆ PRS – âˆ PQR = âˆ QPR,

â‡’ âˆ QTR = Â½ âˆ QPR

Hence, Proved.

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