Lines and Angles

Exercise 6.1

Write the correct answer in each of the following:

1. In Fig. 6.1, if AB || CD || EF, PQ || RS, ∠RQD= 25° and ∠CQP = 60°, then ∠QRS is equal
to
(A) 85°
(B) 135°
(C) 145°
(D) 110°

Sol. We have PQ||RS. Produce PQ to M.

∠CQP = ∠MQD [Vertically opp. ∠s ]
∴ 600 = ∠1 + 250
⇒ ∠1 = 350
Now, QM||RS and QR cuts them.
∠ARQ=∠RQD=250 [Alt. ∠s ]
∴ ∠1 + (∠ARQ+∠ARS)=1800
⇒ 350 (250 + ∠ARS)=1800
⇒ ∠ARS=1800 − 600 = 1200
∴ ∠QRS=∠ARQ+∠ARS=250 + 1200 = 1450
Hence, (c) is the correct answer.

2. If one angle of a triangle is equal to the sum of the other two angles, then the
triangle is
(A) an isosceles triangle
(B) an obtuse triangle
(C) an equilateral triangle
(D) a right triangle
Sol. Let the angles of ∆ABC be ∠A, ∠B and ∠C
Given that ∠A= ∠B+∠C …(1)
 But, in any ∆ABC, ∠A+∠B+∠C=1800 …(2)
[Angles sum property of triangle]
From equations (1) and (2), we get
∠A+∠A=1800 ⇒ 2∠A=1800 ⇒ ∠A=1800 / 2 = 900
∴ A = 900
Hence, the triangle is a right triangle and option (d) is correct.

3. An exterior angle of a triangle is 105° and its two interior opposite angles are
equal. Each of these equal angles is
10
(a) 37
2
10
(b) 52
2
10
(c) 72
2
0
(d) 75
Sol. An exterior angle of triangle is 1500.
Let each of the two interior opposite angles be x.
We know that exterior angle of a triangle is equal to the sum of two interior opposite
angles.
∴ 1500 = x + x ⇒ 2 x = 1500
1 0 10
⇒ x = ×150 = 52
2 2
10
So, each of equal angle is 52
2
Hence, (b) is the correct answer.

4. The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is

(A) an acute angled triangle
(B) an obtuse angled triangle
(C) a right triangle
(D) an isosceles triangle
Sol. Let the angles of the triangle be 5x, 3x and 7x.
As the sum of the angles of a triangle is 1800 , then
5 x + 3x + 7 x = 1800
⇒ 15 x = 1800 ⇒ x = 1800 ÷ 15 = 120
Therefore, the angle of the triangle are
5 × 120 ,3 ×120 and 7 × 120 , i.e., 600 ,360 and 840
As the measure of each angle of the triangle is less than 900 , so the angles of tangle are
acute angles.
Therefore, the triangle is an acute angled triangle.
 Hence, (a) is the correct answer.

5. If one of the angles of a triangle is 130°, then the angle between the bisectors of the
other two angles can be
(A) 50°
(B) 65°
(C) 145°
(D) 155°
Sol. In ∆ABC, we have ∠A=1300 .
OB and OC are the bisectors of the angles B and C.
Now, ∠BOC=1800 − ( ∠OBC+∠OCB)
= 1800 − 250 = 1550
Hence, (d) is the correct answer.

6. In Fig. 6.2, POQ is a line. The value of x is

(A) 20°
(B) 25°
(C) 30°
(D) 35°

Sol. We have 3x + 4 x + 400 = 1800

⇒ 7 x + 400 = 1800 ⇒ 7 x = 1800 − 400 = 1400
⇒ x = 1400 ÷ 7 = 200
Hence, (a) is the correct answer.

7. In Fig. 6.3, if OP||RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR is equal to
(A) 40°
(B) 50°
(C) 60°
(D) 70°

Sol. In the given figure, producing OP, which intersect RQ at X.

 Since, OP||RS and RS is a transversal.
So, ∠RXP=∠XRS [Alternate angles]
0
⇒ ∠RXP=130 …(1) [∵ ∠QRS=1300 ]

Now, RQ is a line segment.

So, ∠PXQ+∠RXP=1800
⇒ ∠PXQ=1800 − ∠RXP=1800 − 1300 [From equation (1)]
⇒ ∠PXQ=500
In ∆PQX,∠OPQ is an exterior angle.
∴ ∠OPQ=∠PXQ+∠PQX
[∵ Exterior angle = sum of two opposite interior angles]
⇒ 110 = 500 + ∠PQX
0

⇒ ∠PQX=1100 − 500
⇒ ∠PQX=600
∴ ∠PQR=600 [∵ ∠PQX=∠PQR]
Hence, the option (c) is correct.

8. Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
(A) 60°
(B) 40°
(C) 80°
(D) 20°
Sol. Given that: The ratio of angles of a triangle is 2 : 4 : 3.
Let the angles of the triangle be ∠A, ∠B and ∠C
∴∠A = 2 x,∠B = 4 x and ∠C = 3 x
In ∠ABC , ∠A + ∠B + ∠C = 1800
[∵ Sum of angles of a triangle is 1800 ]
⇒ 2 x + 4 x + 3x = 1800 ⇒ 9 x = 1800 ⇒ x = 1800 / 9 = 200
∴∠A = 2 x = 2 × 200 = 400
∠B = 4 x = 4 × 200 = 800
And ∠C = 3x = 3 × 200 = 600
Hence, the smallest angles of a triangle is 400 and option (b) is correct answer.
 Lines and Angles
Exercise 6.2

1. For what value of x + y in Fig. 6.4 will ABC be a line? Justify your answer.

Sol. In the given figure, x and y are two adjacent angles.

For ABC to be a straight line, the sum of two adjacent angles x and y must be 1800.

2. Can a triangle have all angles less than 60°? Give reason for your answer.
Sol. A triangle cannot have all angle less than 60°. Then, sum of all the angles will be less than
180° whereas sum of all the angles of a triangle is always 180°.

3. Can a triangle have two obtuse angles? Give reason for your answer.
Sol. An angle whose measure is more than 90o but less than 180o is called an obtuse angle.
A triangle cannot have two obtuse angles because the sum of all the angles of it cannot be
more than 180o. It is always equal to 180o.

4. How many triangles can be drawn having its angles as 45°, 64° and 72°? Give
reason for your answer.
Sol. We cannot draw any triangle having its angles 45°, 64° and 72° because the sum of the
angles (45° + 64° + 72° =181°) cannot be 181°.

5. How many triangles can be drawn having is angles as 53°, 64° and 63°? Give reason
for your answer.
Sol. Sum of these angles = 53° + 64° + 63° = 180°. So, we can draw infinitely many triangles,
sum of the angles of every triangle having its angles as 53°, 64° and 63° is 180°.

6. In Fig. 6.5, find the value of x for which the lines l and m are parallel.
Sol. If a transversal intersects two parallel lines, then each pair of consecutive interior angles
are supplementary. Here, the two given lines l and m are parallel.
Angles x and 44o, are consecutive interior angles on the same side of the transversal.
Therefore, x + 440 = 1800
Hence, x = 1800 − 440 = 1360

7. Two adjacent angles are equal. Is it necessary that each of these angles will be a
right angle? Justify your answer.
Sol. No, each of these angles will be a right angle only when they form a linear pair, i.e., when
the non-common arms of the given two adjacent angles are two opposite rays.

8. If one of the angles formed by two intersecting lines is a right angle, what can you
say about the other three angles? Give reason for your answer.
Sol. If two intersect each other at a point, then four angles are formed. If one of these four
angles is a right angle, then each of the other three angles will also be a right by linear
pair axiom.

9. In Fig.6.6, which of the two lines are parallel and why?

Sol. For fig(i), a transversal intersects two lines such that the sum of interior angles on the
same side on the same side of the transversal is 132o + 48o = 180o.
Therefore, the line l and m are parallel.
For fig. (ii), a transversal intersects two line such that the sum of interior angles on the
same sides of the transversal is 73o + 106o = 179o.
Therefore, the lines p and q are not parallel.

10. Two lines l and m are perpendicular to the same line n. Are l and m perpendicular
to each other? Give reason for your answer.
Sol. When two lines l and m are perpendicular to the same line n, each of the two
corresponding angles formed by these lines l and m with the line n are equal (each is
equal to 90o). Hence, the line l and m are parallel.
 Lines and Angles
Exercise 6.3

1. In Fig. 6.9, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show
that the points A, O and B are collinear.

Sol. Given: In figure, OD ⊥ OE, OD and OE are the bisector of ∠AOC and ∠BOC.
To prove: points A, O and B are collinear i.e., AOB is a straight line.
Proof: Since, OD and OE bisect angles ∠AOC and ∠BOC respectively.
∴ ∠AOC = 2∠DOC …(1)
And ∠COB = 2∠COE …(2)
On adding equations (1) and (2), we get
∠AOC = ∠COB = 2∠DOC + 2∠COE
⇒ ∠AOC + ∠COB = 2(∠DOC + ∠COE )
⇒ ∠AOC + ∠COB = 2∠DOE
⇒ ∠AOC + ∠COB = 2 × 900 [∵ OD ⊥ OE ]
⇒ ∠AOC + ∠COB = 1800
∴ ∠AOB = 1800
So, ∠AOC + ∠COB are forming linear pair or AOB is a straight line. Hence, points A, O
and B are collinear.

2. In Fig. 6.10, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

Sol. We have,
∠5 + ∠6 = 1800 [Angles pf a linear pair]
⇒ ∠5 + 120 = 180 ⇒ ∠5 = 1800 − 1200 = 600
0 0
 Now, ∠1 = ∠5 [Each = 60o]
But, these are corresponding angles.
Therefore, the lines m and n are parallel.

3. AP and BQ are the bisectors of the two alternate interior angles formed by
intersection of a transversal t with parallel lines l and m (Fig. 6.1 1). Show that AP ||
BQ.

Sol. ∵ l || m and t is the transversal

∠MAB = ∠SBA
[Alt. ∠s ]
1 1
⇒ ∠MAB = ∠SBA ⇒ ∠2 = ∠3
2 2
But, ∠2 and ∠3 are alternate angles.
Hence, AP||BQ.

4. If in Fig. 6.1 1, bisectors AP and BQ of the alternate interior angles are parallel, then
show that l ||m.
Sol.

AP is the bisector of ∠MAB and BQ is the bisector of ∠SBA. We are given that AP||BQ.
As AP||BQ, So ∠2 = ∠3 [Alt. ∠s ]
∴ 2 ∠ 2 = 2 ∠3
⇒ ∠ 2 + ∠2 = ∠3 + ∠3
⇒ ∠1 + ∠2 = ∠3 + ∠ 4 [∵ ∠1 = ∠2and ∠3 = ∠4]
⇒ ∠MAB = ∠SBA
 But, these are alternate angles. Hence, the lines l and m are parallel, i.e., l ||m.

5. In Fig. 6.12, BA|| ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to
intersect BC at P (say)].

Sol. Produce DE to intersect BC at P(say).

EF||BC and DP is the transversal,

∴ ∠DEF = ∠DPC …(1) [Corres. ∠s ]

Now, AB||DP and BC is the transversal,
∴ ∠DPC = ∠ABC …(2) [Corres. ∠s ]
From (1) and (2), we get
∠ABC = ∠DEF
Hence, Proved.

6. In Fig. 6.13, BA || ED and BC || EF. Show that ∠ ABC + ∠ DEF = 180°

Sol. Produce ED to meet BC at P(say)

Now, EF||BC and EP is the transversal.

∴ ∠DEF = ∠EPC = 1800 …(1)
Again, EP||AB and BC is the transversal.
∴ ∠EPC = ∠ABC …(2)[corresponding ∠s ]
From (1) and (2), we get
∠DEF = ∠ABC = 1800
⇒ ∠ABC + ∠DEF = 1800
Hence, proved.

7. In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively.
Find ∠APB.

Sol. DE||QR and the line n is the transversal line.

∴∠EAB + ∠RBA = 1800 …(1)
[∵ If a transversal intersects two parallel lines, then each pair of consecutive
interior angles are supplementary]
⇒ ∠PAB + ∠PBA = 900
[∵ AP is the bisector of ∠EAB and BP is the bisector of ∠RBA ]
Now, from ∆APB, we have
∠APB = 1800 − ( ∠PAB + ∠PBA)
⇒ ∠APB = 1800 − 900 = 900
8. The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
Sol. Given: Ratio of angles is 2 : 3 : 4.
To find: Angles of triangle.
Proof: The ratio of angles of a triangle is 2 : 3 : 4.
Let the angles of a triangle be ∠A, ∠B and ∠C
Therefore, ∠A = 2 x, then ∠B = 3 x and ∠C = 4 x.
In ∆ABC , ∠A + ∠B + ∠C = 1800 [∵ Sum of angles of a triangle is 180o]
∴ 2 x + 3x + 4 x = 1800
⇒ 9 x = 1800 ⇒ x = 1800 / 9 = 200
∴ ∠A = 2 x = 2 × 200 = 400
∠B = 3x = 3 × 200 = 600
And ∠C = 4 x = 4 × 200 = 800
Hence, the angles of the triangles are 40o, 60o and 80o.

9. A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that
∠BAL = ∠ACB.
Sol.

Given: In ∆ABC
∠A = 900 and AL ⊥ BC .
To prove: ∠BAL = ∠ACB.
Proof: In ∆ABC and ∆LAC ,
∠BAC = ∠ALC …(1) [Each = 90o]
And ∠ABC = ∠ABL. ...(2) [Common angle]
Adding equations (1) and (2), we get
∠BAC + ∠ABC = ∠ALC + ∠ABC …(3)
In ∆ABC , ∠BAC + ∠ACB + ∠ABC = 1800
[Sum of angles of triangle is 180o]
⇒ ∠BAC + ∠ABC = 1800 − ∠ACB …(4)
In ∆ABL, ∠ABL + ∠ALB + ∠BAL = 1800
[Sum of the angles of triangle is 180o]
⇒ ∠ABL + ∠ALC = 1800 − ∠BAL ...(5)  ∠ALC = ∠ALB = 900 
Substituting the value from equation (4) and (5) in equation (3), we get
1800 − ∠ACB = 1800 − ∠BAL ⇒ ∠ACB = ∠BAL
Hence, proved.
10. Two lines are respectively perpendicular to two parallel lines. Show that they are
parallel to each other.
Sol.

Two lines p and n are respectively perpendicular to two parallel line l and m, i.e., p ⊥ l
and n ⊥ m.
We have to show that p is parallel to n.
As n ⊥ m, so ∠1 = 900 …(1)
Again, p ⊥ l , So ∠2 = 900.
But, l is parallel to m, so
∠2 = ∠3 [corres. ∠s ]
0
∴ ∠2 = ∠90 …(2) [∵ ∠2 = 900 ]
From (1) and (2), we get
⇒ ∠1 = ∠3 [Each = 90o]
But, these are corresponding angles.
Hence, p||n.
 Lines and Angles
Exercise 6.4

1. If two lines intersect, prove that the vertically opposite angles are equal.
Sol.

Given: Two lines AB and CD intersect at point O.

To prove: (i) ∠AOC = ∠BOD
(ii) ∠AOD = ∠ BOC
Proof: (i) Since, ray OA stands on line CD.
∴ ∠AOC = ∠AOD = 1800 …(1)
[Linear pair axiom]
Similarly, ray OD stands on line AB.
∴ ∠AOD = ∠BOD = 1800 …(2)
From equations (1) and (2), we get
∠AOC = ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD
Hence, proved.

(ii) Since, ray OD stands on line AB.

∴ ∠AOD + ∠BOD = 1800 …(3) [Linear pair axiom]
Similarly, ray OB stands on line CD.
∴ ∠DOB + ∠BOC = 1800 …(4)
From equations (3) and (4), we get
∠AOD + ∠BOD = ∠DOB + ∠BOC ⇒ ∠AOD = ∠BOC
Hence, proved.

2. Bisectors of interior ∠B and exterior ∠ACD of a ∆ABC intersect at the point T. Prove
that
1
∠BTC = ∠BAC
2
Sol. Given: ∆ABC , produce BC to D and the bisectors of ∠ABC and ∠ACD meet at point T.
1
To prove: ∠BTC = ∠BAC
2
 Proof: In ∆ABC , ∠ACD is an exterior angle.
∴ ∠ACD = ∠ABC + ∠CAB
[Exterior angle of a triangle is equal to the sum of two opposite angles]
1 1 1
⇒ ∠ACD = ∠CAB + ∠ABC [Dividing both sides by 2]
2 2 2
1 1
⇒ ∠TCD = ∠CAB + ∠ABC …(1)
2 2
1
[∵ CT is a bisector of ∠ACD ⇒ ∠ACD = ∠TCD ]
2
In ∆BTC , ∠TCD = ∠BTC + ∠CBT
[Exterior angle of the triangle is equal to the sum of two opposite angles]
1
⇒ ∠TCD = ∠BTC + ∠ABC …(2)
2
1
[∵ BT is bisector of ∆ABC ⇒ ∠CBT = ∠ABC ]
2
From equation (1) and (2), we get
1 1 1
∠CAB + ∠ABC = ∠BTC + ∠ABC
2 2 2
1 1
⇒ ∠CAB = ∠BTC or ∠BAC = ∠BTC
2 2
Hence, proved.

3. A transversal intersects two parallel lines. Prove that the bisectors of any pair of
corresponding angles so formed are parallel.
Sol. Given: Two lines DE and QR are parallel and are intersected by transversal at A and B
respectively. Also, BP and AF are the bisector of angles ∠ABR and ∠CAE respectively.
 To prove: EP||FQ
Proof: Given, DE || QR ⇒ ∠CAE = ∠ABR [Corresponding angles]
1 1
⇒ ∠CAE = ∠ABR [Dividing both sides by 2]
2 2
⇒ ∠CAE = ∠ABP
[∵ BP and AF are the bisector of angles ∠ABR and ∠CAE
respectively.
As these are the corresponding angles on the transversal line n and are equal.
Here, EP||FQ.

4. Prove that through a given point, we can draw only one perpendicular to a given
line.
[Hint: Use proof by contradiction].
Sol. From the point P, a perpendicular PM is drawn to the given line AB.
∴ ∠PMB = 900
Let if possible, we can draw another perpendicular PN to the line AB. Then,
∠PMB = 900
∴ ∠PMB = ∠PNB, which is possible only when PM and PN coincide with each other.

Hence, through a given point, we can draw only one perpendicular to a given line.

5. Prove that two lines that are respectively perpendicular to two intersecting lines
intersect each other.
[Hint: Use proof by contradiction].
Sol. Given: Let lines l and m are two intersecting lines. Again, let n and p be another two lines
which are perpendicular to the intersecting lines meet at point D.
To prove: Two lines n and p intersecting at a point.
Proof: Let us consider lines n and p are not intersecting, then it means they are parallel to
each other i.e., n||P. …(1)
Since, lines n and p are perpendicular to m and l respectively.
But from equation (1), n||p, it implies that l and m. It is a contradiction
Thus, our assumption is wrong. Hence, lines n and p intersect at a point.

6. Prove that a triangle must have at least two acute angles.

Sol. If the triangle is an acute angled triangle, then all its three angles are acute angle. Each of
these angles is less than 90o, so they can make three angles sum equal to 180o.
If a triangle is a right triangle, then one angle which is right angle will be equal to 90o and
the other two acute angles can make the three angles sum equal to 180o.
 Hence, we can say that a triangle must have a least two acute angles.

7. In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that
1
∠APM = ( ∠Q − ∠R)
2

Sol. Given: ∆PQR, ∠Q < ∠R, PA is the bisector of ∠QPR and PM ⊥ QR.
1
To prove: ∠APM = (∠Q − ∠R)
2
Proof: Since, PA is the bisector of ∠QPR
∠QPA = ∠APR
In ∠PQM , ∠Q + ∠PMQ + ∠QPM = 1800 …(1)
[Angle sum property of a triangle]
⇒ ∠Q + 90 + ∠QPM = 1800
0
[∴∠PMR = 900 ]
⇒ ∠Q = 900 − ∠QPM …(2)
In ∆PMR, ∠PMR + ∠R + ∠RPM = 1800
[Angle sum property of a triangle]
⇒ 90 + ∠R + ∠RPM = 1800
0
[∵ ∠PMR = 900 ]
⇒ ∠R = 1800 − 900 − ∠RPM
⇒ ∠Q = 900 − ∠QPM
⇒ ∠PRM = 900 − ∠RPM …(3)
Subtracting equation (3) from equation (2), we get
∠Q − ∠R = (900 − ∠QPM ) − (900 − ∠RPM )
⇒ ∠Q − ∠R = ∠RPM − ∠QPM
⇒ ∠Q − ∠R = (∠RPM + ∠APM ) −(∠QPA − ∠APM ) …(4)
⇒ ∠Q − ∠R = ∠QPA + ∠APM −∠QPA + ∠APM [Using equation (1)]
⇒ ∠Q − ∠R = 2∠APM
1
⇒ ∠APM = (∠Q − ∠R)
2
Hence, proved.