Lines and Angles – Exercise 3

Question 1: In Figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:

Given: ∠TQP = 110°, ∠SPR = 135°

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 

⇒ ∠TQP + ∠PQR = 180°  ———-(i)

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

⇒ 110° + ∠PQR = 180° 

⇒ ∠PQR = 70°

Consider the ΔPQR,

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

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

Or, ∠PQR + ∠PRQ = 135° ———(ii)

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

∠PRQ = 135° – 70°

Hence, ∠PRQ = 65°

Question 2: In Figure, ∠X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ.

Solution:

Given: ∠X = 62°, ∠XYZ = 54°

As we have studied in this chapter,

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

So, ∠X +∠XYZ +∠XZY = 180°

Putting the values as given in the question we get,

62°+54° + ∠XZY = 180°

Or, ∠XZY = 64°

Now, we know that ZO is the bisector so,

∠OZY = ½ XZY

∴ ∠OZY = 32°

Similarly, YO is a bisector and so,

∠OYZ = ½ XYZ

Or, ∠OYZ = 27° (As XYZ = 54°)

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

∠OZY +∠OYZ +O = 180°

Putting their respective values, we get,

∠O = 180°-32°-27°

Hence, ∠O = 121°

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

Solution:

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

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

Here, BAC and AED are alternate interior angles.

Hence, ∠BAC = ∠AED

 ∠BAC = 35° (Given)

∠AED = 35°

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

∴ ∠DCE + ∠CED + ∠CDE = 180°

Putting the values, we get

∠DCE + 35° + 53° = 180°

Hence, ∠DCE = 92°

Question 4: In Figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:

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

In △PRT.

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

⇒ ∠PTR = 45°

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

⇒ ∠PTR = ∠STQ = 45°

Again, in triangle STQ,

⇒ ∠TSQ +∠PTR + ∠SQT = 180°  (The Sum of all the angles of Triangle is 180°)

Solving this we get,

⇒ ∠SQT = 60°

Question 5: In Figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:

Given: PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°

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

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

⇒ x + 28° = 65°

∴ x = 37°

It is also known that alternate interior angles are same and 

⇒ QSR = x = 37°

Also, 

⇒ QRS + QRT = 180° (As they form a Linear pair)

Putting the value of ∠QRT = 65° we get,

⇒ QRS + 65° = 180°

⇒ QRS = 115°

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

⇒ P + Q + R + S = 360°

Putting their respective values, we get,

⇒ S = 360° – 90° – 65° – 115° = 900

In Δ SPQ

⇒ ∠SPQ + x + y = 1800

⇒ 900 + 370 + y = 1800

⇒ y = 1800 – 1270 = 530

Hence, y = 53°

Question 6: In 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.

Solution:

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

To Prove: ∠QTR = ½ ∠QPR

Proof:

Consider the ΔPQR. 

∠PRS is an exterior angle.

∠QPR and ∠PQR are interior angles.

⇒ ∠PRS = ∠QPR + ∠PQR (According to triangle property)

⇒ ∠PRS – ∠PQR = ∠QPR  ————(i)

Now, consider the ΔQRT,

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

⇒ ∠QTR = ∠TRS – ∠TQR

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

So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR

Now,

⇒ ∠QTR = ½ ∠PRS – ½ ∠PQR

⇒ ∠QTR = ½ ∠(PRS – PQR)

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

⇒ ∠QTR = ½ ∠QPR 

Hence, Proved.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles is curated and compiled by the expert team of professionals at GFG to assist students in resolving questions related to lines and angles they may have as they go through problems from the NCERT textbook.

This chapter Lines and Angles mainly covers the characteristics of the angles created when two or more lines cross one another. Additionally, it gives a fundamental comprehension of the definitions of various terminology used in geometry. The many kinds of angles, pairs of angles, transversals, parallel lines, the angle sum property of triangles, and axioms related to these ideas are all taught to students.