Angles are formed when two rays start from the same point. The rays forming an angle are known as arm and the end point is the vertex of the angle. There are different kinds of angles namely reflex angle, obtuse angle, right angle, complete angle, straight angle and acute angle. The different types of lines mainly include skew lines, vertical lines, parallel lines, horizontal lines and perpendicular lines.

RS Aggarwal Solutions for Class 9 helps the students to score good marks in the board exams. The problems are solved accurately with step by step solutions according to the RS Aggarwal Maths text book. RS Aggarwal Solutions for Class 9 Chapter 7 Lines and Angles are provided here.

## RS Aggarwal Solutions for Class 9 Maths Chapter 7 : Lines and Angles Download PDF

## Access RS Aggarwal Solutions for Class 9 Maths Chapter 7: Lines and Angles

Exercise 7(A) PAGE: 198

**1. Define the following terms:**

**(i) Angle**

**(ii) Interior of an angle**

**(iii) Obtuse angle**

**(iv) Reflex angle**

**(v) Complementary angles**

**(vi) Supplementary angles**

**Solution:**

(i) Angle â€“ When two rays originate from the same end point, then an angle is formed.

(ii) Interior of an angle â€“ The interior of âˆ BAC is the set of all points in its plane which lie on the same side of AB as C and also on the same side of AC as B.

(iii)Obtuse angle â€“ An angle whose measure is more than 90^{o} but less than 180^{o} is called an obtuse angle.

(iv) Reflex angle â€“ An angle whose measure is more than 180^{o} but less than 360^{o} is called a reflex angle.

(v) Complementary angles â€“ Two angles are said to be complementary, if the sum of their measures is 90^{o}.

(vi) Supplementary angles â€“ Two angles are said to be supplementary if the sum of their measures is 180^{o}.

**2. Find the complement of each of the following angles:**

**(i) 55 ^{o}**

**(ii) 16 ^{o}**

**(iii) 90 ^{o}**

**(iv) 2/3 of a right angle**

**Solution:**

(i) We know that the complement of 55^{o} can be written as

55^{o} = 90^{o} – 55^{o} = 35^{o}

(ii) We know that the complement of 16^{o} can be written as

16^{o} = 90^{o} – 16^{o} = 74^{o}

(iii) We know that the complement of 90^{o} can be written as

90^{o} = 90^{o} – 90^{o} = 0^{o}

(iv) We know that 2/3 of a right angle can be written as 2/3 Ã— 90^{o} = 60^{o}

60^{o }= 90^{o} – 60^{o} = 30^{o}

**3. Find the supplement of each of the following angles:**

**(i) 42 ^{o}**

**(ii) 90 ^{o}**

**(iii) 124 ^{o}**

**(iv) 3/5 of a right angle**

**Solution:**

(i) We know that the supplement of 42^{o} can be written as

42^{o} = 180^{o} – 42^{o} = 138^{o}

(ii) We know that the supplement of 90^{o} can be written as

90^{o} = 180^{o} – 90^{o} = 90^{o}

(iii) We know that the supplement of 124^{o} can be written as

124^{o} = 180^{o} – 124^{o} = 56^{o}

(iv) We know that 3/5 of a right angle can be written as 3/5 Ã— 90^{o} = 54^{o}

54^{o }= 180^{o} – 54^{o} = 126^{o}

**4. Find the measure of an angle which is**

**(i) Equal to its complement**

**(ii) Equal to its supplement**

**Solution:**

(i) Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o}

To find that the measure of an angle is equal to its complement

We get

x^{o} = 90^{o} – x^{o}

We can also write it as

x + x = 90

So we get

2x = 90

By division we get

x = 90/2

x^{o} = 45^{o}

Therefore, the measure of an angle which is equal to its complement is 45^{o}

(ii) Consider the required angle as x^{o}

We know that the supplement can be written as 180^{o} – x^{o}

To find that the measure of an angle is equal to its supplement

We get

x^{o} = 180^{o} – x^{o}

We can also write it as

x + x = 180

So we get

2x = 180

By division we get

x = 180/2

x^{o} = 90^{o}

Therefore, the measure of an angle which is equal to its supplement is 90^{o}

**5. Find the measure of an angle which is 36 ^{o} more than its complement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o}

x^{o} = (90^{o} – x^{o}) + 36^{o}

We can also write it as

x + x = 90 + 36

So we get

2x = 126

By division we get

x = 126/2

x^{o} = 63^{o}

Therefore, the measure of an angle which is 36^{o} more than its complement is 63^{o}.

**6. Find the measure of an angle which is 30 ^{o} less than its supplement.**

**Solution:**

Consider the required angle as x^{o}

We know that the supplement can be written as 180^{o} – x^{o}

x^{o} = (180^{o} – x^{o}) â€“ 30^{o}

We can also write it as

x + x = 180 – 30

So we get

2x = 150

By division we get

x = 150/2

x^{o} = 75^{o}

Therefore, the measure of an angle which is 30^{o} less than its supplement is 75^{o}.

**7. Find the angle which is four times its complement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o}

x^{o} = 4(90^{o} – x^{o})

We can also write it as

x = 360 â€“ 4x

So we get

5x = 360

By division we get

x = 360/5

x^{o} = 72^{o}

Therefore, the angle which is four times its complement is 72^{o}.

**8. Find the angle which is five times its supplement.**

**Solution:**

Consider the required angle as x^{o}

We know that the supplement can be written as 180^{o} – x^{o}

x^{o} = 5(180^{o} – x^{o})

We can also write it as

x = 900 â€“ 5x

So we get

6x = 900

By division we get

x = 900/6

x^{o} = 150^{o}

Therefore, the angle which is five times its supplement is 150^{o}.

**9. Find the angle whose supplement is four times its complement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o} and the supplement can be written as 180^{o} – x^{o}

180^{o} – x^{o} = 4(90^{o} – x^{o})

We can also write it as

180^{o} – x^{o} = 360^{o} – 4x^{o}

So we get

4x^{o} – x^{o} = 360^{o} – 180^{o}

3x^{o} = 180^{o}

By division we get

x^{o} = 180/3

x^{o} = 60^{o}

Therefore, the angle whose supplement is four times its complement is 60^{o}.

**10. Find the angle whose complement is one third of its supplement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o} and the supplement can be written as 180^{o} – x^{o}

90^{o} – x^{o} = 1/3(180^{o} – x^{o})

We can also write it as

90^{o} – x^{o} = 60^{o} â€“ (1/3) x^{o}

So we get

x^{o} â€“ (1/3)x^{o} = 90^{o} – 60^{o}

(2/3) x^{o} = 30^{o}

By division we get

x^{o} = ((30Ã—3)/2)

x^{o} = 45^{o}

Therefore, the angle whose complement is one third of its supplement is 45^{o}.

**11. Two complementary angles are in the ratio 4:5. Find the angles.**

**Solution:**

Consider the required angle as x^{o} and 90^{o} – x^{o}

According to the question it can be written as

x^{o}/ 90^{o} – x^{o} = 4/5

By cross multiplication we get

5x = 4 (90 â€“ x)

5x = 360 â€“ 4x

On further calculation we get

5x + 4x = 360

9x = 360

By division

x = 360/9

So we get

x = 40

Therefore, the angles are 40^{o} and 90^{o} – x^{o} = 90^{o} – 40^{o} = 50^{o}

**12. Find the value of x for which the angles (2x â€“ 5) ^{ o} and (x â€“ 10)^{ o} are the complementary angles.**

**Solution:**

It is given that (2x â€“ 5)^{ o} and (x â€“ 10)^{ o} are the complementary angles.

So we can write it as

(2x â€“ 5)^{ o} + (x â€“ 10)^{ o} = 90^{o}

2x â€“ 5^{o} + x â€“ 10^{o} = 90^{o}

On further calculation

3x – 15^{o} = 90^{o}

So we get

3x = 105^{o}

By division

x = 105/3

x = 35^{o}

Therefore, the value of x for which the angles (2x â€“ 5)^{ o} and (x â€“ 10)^{ o} are the complementary angles is 35^{o}.

Exercise 7(B) PAGE: 206

**1. In the adjoining figure, AOB is a straight line. Find the value of x.**

**Solution:**

From the figure we know that âˆ AOC and âˆ BOC are a linear pair of angles

So we get

âˆ AOC + âˆ BOC = 180^{o }

We know that

62^{o} + x^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 62^{o}

By subtraction

x^{o} = 118^{o}

Therefore, the value of x is 118.

**2. In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find âˆ AOC and âˆ BOD.**

**Solution:**

From the figure we know that âˆ AOB is a straight line

So we get

âˆ AOB = 180^{o}

It can also be written as

âˆ AOC + âˆ COD + âˆ BOD = 180^{o}

By substituting the values

(3x â€“ 7) ^{o} + 55^{o} + (x + 20) ^{o} = 180^{o}

3x â€“ 7^{o} + 55^{o} + x + 20^{o} = 180^{o}

On further calculation

4x + 68^{o} = 180^{o}

4x = 112^{o}

By division

x = 28^{o}

By substituting the value of x we get

âˆ AOC = (3x â€“ 7) ^{o}

= 3(28^{o}) â€“ 7^{o}

On further calculation

= 84^{o} – 7^{o}

By subtraction

= 77^{o}

âˆ BOD = (x + 20) ^{o}

= (28 + 20) ^{o}

By addition

= 48^{o}

**3. In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find âˆ AOC, âˆ COD and âˆ BOD.**

**Solution:**

From the figure we know that âˆ BOD and âˆ AOD are a linear pair of angles

So we get

âˆ BOD + âˆ AOD = 180^{o}

It can also be written as

âˆ BOD + âˆ COD + âˆ COA = 180^{o}

By substituting the values

x^{o} + (2x â€“ 19) ^{o} + (3x + 7) ^{o} = 180^{o}

x + 2x â€“ 19^{o} + 3x + 7^{o} = 180^{o}

On further calculation

6x â€“ 12^{o} = 180^{o}

6x = 180^{o} + 12^{o}

So we get

6x = 192^{o}

By division

x = 32^{o}

By substituting the value of x we get

âˆ AOC = (3x + 7) ^{o}

= 3(32^{o}) + 7^{o}

On further calculation

= 96^{o} + 7^{o}

By addition

= 103^{o}

âˆ COD = (2x – 19) ^{o}

= (2(32^{o}) – 19) ^{o}

So we get

= (64 â€“ 19) ^{o}

By subtraction

= 45^{o}

âˆ BOD = x^{o} = 32^{o}

**4. In the adjoining figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.**

**Solution:**

From the figure it is given that

x: y: z = 5: 4: 6

We can also write it as

x + y + z = 5 + 4 + 6 = 15

It is given that XOY is a straight line

So we know that

x + y + z = 180^{o}

As we know the sum of ratios is 15 then we can write that the measure of x as 5

The sum of all the angles in a straight line is 180^{o}

So we get the measure of x as

x = (5/15) Ã— 180

On further calculation

x = 60

As we know the sum of ratios is 15 then we can write that the measure of y as 4

The sum of all the angles in a straight line is 180^{o}

So we get the measure of y as

y = (4/15) Ã— 180

On further calculation

y = 48

In order to find the value of z

We know that

x + y + z = 180^{o}

Substituting the values of x and y we get

60^{o} + 48^{o} + z = 180^{o}

On further calculation

z = 180^{o} – 60^{o} – 48^{o}

By subtraction we get

z = 72^{o}

Therefore, the values of x, y and z are 60^{o}, 48^{o} and 72^{o}.

**5. In the adjoining figure, what value of x will make AOB, a straight line?**

**Solution:**

We know that AOB will be a straight line only if the adjacent angles form a linear pair.

âˆ BOC + âˆ AOC = 180^{o}

By substituting the values we get

(4x â€“ 36) ^{o} + (3x + 20) ^{o} = 180^{o}

4x – 36^{o} + 3x + 20^{o} = 180^{o}

On further calculation we get

7x = 180^{o} – 20^{o} + 36^{o}

7x = 196^{o}

By division we get

x = 196/7

x = 28

Therefore, the value of x is 28.

**6. Two lines AB and CD intersect at O. If âˆ AOC = 50 ^{o}, find âˆ AOD, âˆ BOD and âˆ BOC.**

**Solution:**

From the figure we know that âˆ AOC and âˆ AOD form a linear pair.

It can also be written as

âˆ AOC + âˆ AOD = 180^{o}

By substituting the values

50^{o} + âˆ AOD = 180^{o}

âˆ AOD = 180^{o} – 50^{o}

By subtraction

âˆ AOD = 130^{o}

According to the figure we know that âˆ AOD and âˆ BOC are vertically opposite angles

So we get

âˆ AOD = âˆ BOC = 130^{o}

According to the figure we know that âˆ AOC and âˆ BOD are vertically opposite angles

So we get

âˆ AOC = âˆ BOD = 50^{o}

**7. In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.**

**Solution:**

From the figure we know that âˆ COE and âˆ DOF are vertically opposite angles

âˆ COE = âˆ DOF = âˆ z = 50^{o}

From the figure we know that âˆ BOD and âˆ COA are vertically opposite angles

âˆ BOD = âˆ COA = âˆ t = 90^{o}

We also know that âˆ COA and âˆ AOD form a linear pair

âˆ COA + âˆ AOD = 180^{o}

It can also be written as

âˆ COA + âˆ AOF + âˆ FOD = 180^{o}

By substituting values in the above equation we get

90^{o} + x^{o} + 50^{o} = 180^{o}

On further calculation we get

x^{o} + 140^{o} = 180^{o}

x^{o} = 180^{o} – 140^{o}

By subtraction

x^{o} = 40^{o}

From the figure we know that âˆ EOB and âˆ AOF are vertically opposite angles

âˆ EOB = âˆ AOF = x = y = 40

Therefore, the values of x, y, z and t are 40, 40, 50 and 90.

**8. In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence, find âˆ AOD, âˆ COE and âˆ AOE.**

**Solution:**

From the figure we know that âˆ COE and âˆ EOD form a linear pair

âˆ COE + âˆ EOD = 180^{o}

It can also be written as

âˆ COE + âˆ EOA + âˆ AOD = 180^{o}

By substituting values in the above equation we get

5x + âˆ EOA + 2x = 180^{o}

From the figure we know that âˆ EOA and âˆ BOF are vertically opposite angles

âˆ EOA = âˆ BOF

So we get

5x + âˆ BOF + 2x = 180^{o}

5x + 3x + 2x = 180^{o}

On further calculation

10x = 180^{o}

By division

x = 180/10 = 18

By substituting the value of x

âˆ AOD = 2x^{o}

So we get

âˆ AOD = 2 (18) ^{o} = 36^{o}

âˆ EOA = âˆ BOF = 3x^{o}

So we get

âˆ EOA = âˆ BOF = 3 (18) ^{o} = 54^{o}

âˆ COE = 5x^{o}

So we get

âˆ COE = 5 (18) ^{o} = 90^{o}

**9. Two adjacent angles on a straight line are in the ratio 5:4. Find the measure of each one of these angles.**

**Solution:**

Consider the two adjacent angles as 5x and 4x.

We know that the two adjacent angles form a linear pair

So it can be written as

5x + 4x = 180^{o}

On further calculation

9x = 180^{o}

By division

x = 180/9

x = 20^{o}

Substituting the value of x in two adjacent angles

5x = 5 (20) ^{o} = 100^{o}

4x = 4 (20) ^{o} = 80^{o}

Therefore, the measure of each one of these angles is 100^{o} and 80^{o}

**10. If two straight lines intersect each other in such a way that one of the angles formed measures 90 ^{o}, show that each of the remaining angles measures 90^{o}.**

**Solution:**

Consider two lines AB and CD intersecting at a point O with âˆ AOC = 90^{o}

From the figure we know that âˆ AOC and âˆ BOD are vertically opposite angles

âˆ AOC = âˆ BOD = 90^{o}

From the figure we also know that âˆ AOC and âˆ AOD form a linear pair

So it can be written as

âˆ AOC + âˆ AOD = 180^{o}

Substituting the values

90^{o} + âˆ AOD = 180^{o}

On further calculation

âˆ AOD = 180^{o} – 90^{o}

By subtraction

âˆ AOD = 90^{o}

From the figure we also know that âˆ BOC and âˆ AOD are vertically opposite angles

âˆ BOC = âˆ AOD = 90^{o}

Therefore, it is proved that each of the remaining angles is 90^{o}

**11. Two lines AB and CD intersect at a point O such that âˆ BOC + âˆ AOD = 280 ^{o}, as shown in the figure. Find all the four angles.**

**Solution:**

From the figure we know that âˆ AOD and âˆ BOC are vertically opposite angles

âˆ AOD = âˆ BOC

It is given that

âˆ BOC + âˆ AOD = 280^{o}

We know that âˆ AOD = âˆ BOC

So it can be written as

âˆ AOD + âˆ AOD = 280^{o}

On further calculation

2 âˆ AOD = 280^{o}

By division

âˆ AOD = 280/2

âˆ AOD = âˆ BOC = 140^{o}

From the figure we know that âˆ AOC and âˆ AOD form a linear pair

So it can be written as

âˆ AOC + âˆ AOD = 180^{o}

Substituting the values

âˆ AOC + 140^{o} = 180^{o}

On further calculation

âˆ AOC = 180^{o} – 140^{o}

By subtraction

âˆ AOC = 40^{o}

From the figure we know that âˆ AOC and âˆ BOD are vertically opposite angles

âˆ AOC = âˆ BOD = 40^{o}

Therefore, âˆ AOC = 40^{o}, âˆ BOC = 140^{o}, âˆ AOD = 140^{o} and âˆ BOD = 40^{o}

**12. Two lines AB and CD intersect each other at a point O such that âˆ AOC: âˆ AOD = 5:7. Find all the angles.**

**Solution:**

Consider âˆ AOC as 5x and âˆ AOD as 7x

From the figure we know that âˆ AOC and âˆ AOD for linear pair of angles

So it can be written as

âˆ AOC + âˆ AOD = 180^{o}

By substituting the values

5x + 7x = 180^{o}

On further calculation

12x = 180^{o}

By division

x = 180/12

x = 15^{o}

By substituting the value of x

âˆ AOC = 5x

So we get

âˆ AOC = 5 (15^{o}) = 75^{o}

âˆ AOD = 7x

So we get

âˆ AOD = 7 (15^{o}) = 105^{o}

From the figure we know that âˆ AOC and âˆ BOD are vertical angles

So we get

âˆ AOC = âˆ BOD = 75^{o}

From the figure we know that âˆ AOD and âˆ BOC are vertical angles

So we get

âˆ AOD = âˆ BOC = 105^{o}

**13. In the given figure, three lines AB, CD and EF intersect at a point O such that âˆ AOE = 35 ^{o} and âˆ BOD = 40^{o}. Find the measure of âˆ AOC, âˆ BOF, âˆ COF and âˆ DOE.**

**Solution:**

It is given that âˆ BOD = 40^{o}

From the figure we know that âˆ BOD and âˆ AOC are vertically opposite angles

âˆ AOC = âˆ BOD = 40^{o}

It is given that âˆ AOE = 35^{o}

From the figure we know that âˆ BOF and âˆ AOE are vertically opposite angles

âˆ AOE = âˆ BOF = 35^{o}

From the figure we know that AOB is a straight line

So it can be written as

âˆ AOB = 180^{o}

We can write it as

âˆ AOE + âˆ EOD + âˆ BOD = 180^{o}

By substituting the values

35^{o} + âˆ EOD + 40^{o} = 180^{o}

On further calculation

âˆ EOD = 180^{o} – 35^{o} – 40^{o}

By subtraction

âˆ EOD = 105^{o}

From the figure we know that âˆ COF and âˆ EOD are vertically opposite angles

âˆ COF = âˆ EOD = 105^{o}

**14. In the given figure, the two lines AB and CD intersect at a point O such that âˆ BOC = 125 ^{o}. Find the values of x, y and z.**

**Solution:**

From the figure we know that âˆ AOC and âˆ BOC form a linear pair of angles

So it can be written as

âˆ AOC + âˆ BOC = 180^{o}

By substituting the values we get

x + 125^{o} = 180^{o}

On further calculation

x = 180^{o} – 125^{o}

By subtraction

x = 55^{o}

From the figure we know that âˆ AOD and âˆ BOC are vertically opposite angles

So we get

y = 125^{o}

From the figure we know that âˆ BOD and âˆ AOC are vertically opposite angles

So we get

z = 55^{o}

Therefore, the values of x, y and z are 55^{o}, 125^{o} and 55^{o}

**15. If two straight lines intersect each other than prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.**

**Solution:**

Given: Let us consider AB and CD as the two lines intersecting at a point O where OE is the ray bisecting âˆ BOD and OF is the ray bisecting âˆ AOC.

To Prove: âˆ AOF = âˆ COF

Proof: EF is a straight line passing through the point O where

are two opposite rays.

From the figure we know that âˆ AOF and âˆ BOE, âˆ COF and âˆ DOE are vertically opposite angles

So it can be written as

âˆ AOF = âˆ BOE and âˆ COF = âˆ DOE

It is given that âˆ BOE = âˆ DOE

So we can write it as âˆ AOF = âˆ COF

Therefore, it is proved that âˆ AOF = âˆ COF.

**16. Prove that the bisectors of two adjacent supplementary angles include a right angle.**

**Solution:**

Given:

is the bisector of âˆ ACD and

is the bisector of âˆ BCD

To Prove: âˆ ECF = 90^{o}

Proof:

From the figure we know that

âˆ ACD and âˆ BCD form a linear pair of angles

So we can write it as

âˆ ACD + âˆ BCD = 180^{o}

We can also write it as

âˆ ACE + âˆ ECD + âˆ DCF + âˆ FCB = 180^{o}

From the figure we also know that

âˆ ACE = âˆ ECD and âˆ DCF = âˆ FCB

So it can be written as

âˆ ECD + âˆ ECD + âˆ DCF + âˆ DCF = 180^{o}

On further calculation we get

2 âˆ ECD + 2 âˆ DCF = 180^{o}

Taking out 2 as common we get

2 (âˆ ECD + âˆ DCF) = 180^{o}

By division we get

(âˆ ECD + âˆ DCF) = 180/2

âˆ ECD + âˆ DCF = 90^{o}

Therefore, it is proved that âˆ ECF = 90^{o}

Exercise 7(C) PAGE: 223

**1. In the given figure, l || m and a transversal t cuts them. If âˆ 1 = 120 ^{o}, find the measure of each of the remaining marked angles.**

**Solution:**

It is given that âˆ 1 = 120^{o}

From the figure we know that âˆ 1 and âˆ 2 form a linear pair of angles

So it can be written as

âˆ 1 + âˆ 2 = 180^{o}

By substituting the values

120^{o} + âˆ 2 = 180^{o}

On further calculation

âˆ 2 = 180^{o} – 120^{o}

By subtraction

âˆ 2 = 60^{o}

From the figure we know that âˆ 1 and âˆ 3 are vertically opposite angles

So we get

âˆ 1 = âˆ 3 = 120^{o}

From the figure we know that âˆ 2 and âˆ 4 are vertically opposite angles

So we get

âˆ 2 = âˆ 4 = 60^{o}

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

âˆ 1 = âˆ 5 = 120^{o}

âˆ 2 = âˆ 6 = 60^{o}

âˆ 3 = âˆ 7 = 120^{o}

âˆ 4 = âˆ 8 = 60^{o}

**2. In the figure, l || m and a transversal t cuts them. If âˆ 7 = 80 ^{o}, find the measure of each of the remaining marked angles.**

**Solution:**

It is given that âˆ 7 = 80^{o}

From the figure we know that âˆ 7 and âˆ 8 form a linear pair of angles

So it can be written as

âˆ 7 + âˆ 8 = 180^{o}

By substituting the values

80^{o} + âˆ 8 = 180^{o}

On further calculation

âˆ 8 = 180^{o} – 80^{o}

By subtraction

âˆ 8 = 100^{o}

From the figure we know that âˆ 7 and âˆ 5 are vertically opposite angles

So we get

âˆ 7 = âˆ 5 = 80^{o}

From the figure we know that âˆ 6 and âˆ 8 are vertically opposite angles

So we get

âˆ 6 = âˆ 8 = 100^{o}

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

âˆ 1 = âˆ 5 = 80^{o}

âˆ 2 = âˆ 6 = 100^{o}

âˆ 3 = âˆ 7 = 80^{o}

âˆ 4 = âˆ 8 = 100^{o}

**3. In the figure, l || m and a transversal t cuts them. If âˆ 1: âˆ 2 = 2: 3, find the measure of each of the marked angles.**

**Solution:**

It is given that âˆ 1: âˆ 2 = 2: 3

From the figure we know that âˆ 1 and âˆ 2 form a linear pair of angles

So it can be written as

âˆ 1 + âˆ 2 = 180^{o}

By substituting the values

2x + 3x = 180^{o}

On further calculation

5x = 180^{o}

By division

x = 180^{o}/5

x = 36^{o}

By substituting the value of x we get

âˆ 1 = 2x = 2 (36^{o}) = 72^{o}

âˆ 2 = 3x = 3 (36^{o}) = 108^{o}

From the figure we know that âˆ 1 and âˆ 3 are vertically opposite angles

So we get

âˆ 1 = âˆ 3 = 72^{o}

From the figure we know that âˆ 2 and âˆ 4 are vertically opposite angles

So we get

âˆ 2 = âˆ 4 = 108^{o}

It is given that, l || m and t is a transversal

So the corresponding angles according to the figure is written as

âˆ 1 = âˆ 5 = 72^{o}

âˆ 2 = âˆ 6 = 108^{o}

âˆ 3 = âˆ 7 = 72^{o}

âˆ 4 = âˆ 8 = 108^{o}

**4. For what value of x will the lines l and m be parallel to each other?**

**Solution:**

If the lines l and m are parallel it can be written as

3x â€“ 20 = 2x + 10

We know that the two lines are parallel if the corresponding angles are equal

3x â€“ 2x = 10 + 20

On further calculation

x = 30

Therefore, the value of x is 30.

**5. For what value of x will the lines l and m be parallel to each other?**

**Solution:**

We know that both the angles are consecutive interior angles

So it can be written as

3x + 5 + 4x = 180

On further calculation we get

7x = 180 â€“ 5

By subtraction

7x = 175

By division we get

x = 175/ 7

x = 25

Therefore, the value of x is 25.

**6. In the figure, AB || CD and BC || ED. Find the value of x.**

**Solution:**

From the given figure we know that

AB and CD are parallel line and BC is a transversal

We know that âˆ BCD and âˆ ABC are alternate angles

So we can write it as

âˆ BCD = âˆ ABC = x^{o}

We also know that BC || ED and CD is a transversal

From the figure we know that âˆ BCD and âˆ EDC form a linear pair of angles

So it can be written as

âˆ BCD + âˆ EDC = 180^{o}

By substituting the values we get

âˆ BCD + 75^{o} = 180^{o}

On further calculation we get

âˆ BCD = 180^{o} – 75^{o}

By subtraction

âˆ BCD = 105^{o}

From the figure we know that âˆ BCD and âˆ ABC are alternate angles

So we get

âˆ BCD = âˆ ABC = x = 105^{o}

âˆ ABC = x = 105^{o}

Therefore, the value of x is 105^{o}.

**7. In the figure, AB || CD || EF. Find the value of x.**

**Solution:**

It is given that AB || CD and BC is a transversal

From the figure we know that âˆ BCD and âˆ ABC are alternate interior angles

So we get

âˆ ABC = âˆ BCD

In order to find the value of x we can write it as

x^{o} + âˆ ECD = 70^{o} â€¦â€¦. (1)

It is given that CD || EF and CE is a transversal

From the figure we know that âˆ ECD and âˆ CEF are consecutive interior angles

So we get

âˆ ECD + âˆ CEF = 180^{o}

By substituting the values

âˆ ECD + 130^{o} = 180^{o}

On further calculation we get

âˆ ECD = 180^{o} – 130^{o}

By subtraction

âˆ ECD = 50^{o}

Now by substituting âˆ ECD in equation (1) we get

x^{o} + âˆ ECD = 70^{o}

x^{o} + 50^{o} = 70^{o}

On further calculation we get

x^{o }= 70^{o} – 50^{o}

By subtraction

x^{o }= 20^{o}

Therefore, the value of x is 20^{o}.

**8. In the given figure, AB || CD. Find the values of x, y and z.**

**Solution:**

It is given that AB || CD and EF is a transversal

From the figure we know that âˆ AEF and âˆ EFG are alternate angles

So we get

âˆ AEF = âˆ EFG = 75^{o}

âˆ EFG = y = 75^{o}

From the figure we know that âˆ EFC and âˆ EFG form a linear pair of angles

So we get

âˆ EFC + âˆ EFG = 180^{o}

It can also be written as

x + y = 180^{o}

By substituting the value of y we get

x + 75^{o} = 180^{o}

On further calculation we get

x = 180^{o} – 75^{o}

By subtraction

x = 105^{o}

From the figure based on the exterior angle property it can be written as

âˆ EGD = âˆ EFG + âˆ FEG

By substituting the values in the above equation we get

125^{o} = y + z

125^{o} = 75^{o }+ z

On further calculation we get

z = 125^{o} – 75^{o}

By subtraction

z = 50^{o}

Therefore, the values of x, y and z are 105^{o}, 75^{o} and 50^{o}.

**9. In each of the figures given below, AB || CD. Find the value of x in each case. **

**(i) **

**(ii)**

**(iii)**

**Solution:**

(i) Draw a line at point E which is parallel to CD and name it as EG

It is given that EG || CD and ED is a transversal

From the figure we know that âˆ GED and âˆ EDC are alternate interior angles

So we get

âˆ GED = âˆ EDC = 65^{o}

EG || CD and AB || CD

So we get EG || AB and EB is a transversal

From the figure we know that âˆ BEG and âˆ ABE are alternate interior angles

So we get

âˆ BEG = âˆ ABE = 35^{o}

âˆ DEB = x^{o}

From the figure we can write âˆ DEB as

âˆ DEB = âˆ BEG + âˆ GED

By substituting the values

x^{o} = 35^{o} + 65^{o}

By addition

x^{o} = 100^{o}

Therefore, the value of x is 100.

(ii) Draw a line OF which is parallel to CD

So we get

OF || CD and OD is a transversal

From the figure we know that âˆ CDO and âˆ FOD are consecutive angles

So we get

âˆ CDO + âˆ FOD = 180^{o}

By substituting the values

25^{o} + âˆ FOD = 180^{o}

On further calculation

âˆ FOD = 180^{o} – 25^{o}

By subtraction

âˆ FOD = 155^{o}

We also know that OF || CD and AB || CD

So we get OF || AB and OB is a transversal

From the figure we know that âˆ ABO and âˆ FOB are consecutive angles

So we get

âˆ ABO + âˆ FOB = 180^{o}

By substituting the values

55^{o} + âˆ FOB = 180^{o}

On further calculation

âˆ FOB = 180^{o} – 55^{o}

By subtraction

âˆ FOB = 125^{o}

In order to find the value of x

x^{o} = âˆ FOB + âˆ FOD

By substituting the values

x^{o} = 125^{o} + 155^{o}

By addition

x^{o} = 280^{o}

Therefore, the value of x is 280.

(iii) Draw a line EF which is parallel to CD

So we get EF || CD and EC is a transversal

From the figure we know that âˆ FEC and âˆ ECD are consecutive interior angles

So we get

âˆ FEC + âˆ ECD = 180^{o}

By substituting the values

âˆ FEC + 124^{o} = 180^{o}

On further calculation

âˆ FEC = 180^{o} – 124^{o}

By subtraction

âˆ FEC = 56^{o}

We know that EF || CD and AB || CD

So we get EF || AB and AE is a transversal

From the figure we know that âˆ BAE and âˆ FEA are consecutive interior angles

So we get

âˆ BAE + âˆ FEA = 180^{o}

By substituting the values

116^{o} + âˆ FEA = 180^{o}

On further calculation

âˆ FEA = 180^{o} – 116^{o}

By subtraction

âˆ FEA = 64^{o}

In order to find the value of x

x^{o} = âˆ FEA + âˆ FEC

By substituting the values

x^{o} = 64^{o} + 56^{o}

By addition

x^{o} = 120^{o}

Therefore, the value of x is 120^{o}

**10. In the given figure, AB || CD. Find the value of x.**

**Solution:**

Draw a line through point C and name it as FG where FG || AE

We know that CG || AE and CE is a transversal

From the figure we know that âˆ GCE and âˆ CEA are alternate angles

So we get

âˆ GCE = âˆ CEA = 20^{o}

It can also be written as

âˆ DCG = âˆ DCE – âˆ GCE

By substituting the values we get

âˆ DCG = 130^{o} – 20^{o}

By subtraction we get

âˆ DCG = 110^{o}

We also know that AB || CD and FG is a transversal

From the figure we know that âˆ BFC and âˆ DCG are corresponding angles

So we get

âˆ BFC = âˆ DCG = 110^{o}

We know that FG || AE and AF is a transversal

From the figure we know that âˆ BFG and âˆ FAE are corresponding angles

So we get

âˆ BFG = âˆ FAE = 110^{o}

âˆ FAE = x = 110^{o}

Therefore, the value of x is 110.

**11. In the given figure, AB || PQ. Find the values of x and y.**

**Solution:**

It is given that AB || PQ and EF is a transversal

From the figure we know that âˆ FEB and âˆ EFQ are corresponding angles

So we get

âˆ FEB = âˆ EFQ = 75^{o}

It can be written as

âˆ EFQ = 75^{o}

Where

âˆ EFG + âˆ GFQ = 75^{o}

By substituting the values

25^{o} + y^{o} = 75^{o}

On further calculation

y^{o} = 75^{o} – 25^{o}

By subtraction

y^{o} = 50^{o}

From the figure we know that âˆ BEF and âˆ EFQ are consecutive interior angles

So we get

âˆ BEF + âˆ EFQ = 180^{o}

By substituting the values

âˆ BEF + 75^{o} = 180^{o}

On further calculation

âˆ BEF = 180^{o} – 75^{o}

By subtraction

âˆ BEF = 105^{o}

We know that âˆ BEF can be written as

âˆ BEF = âˆ FEG + âˆ GEB

105^{o} = âˆ FEG + 20^{o}

On further calculation

âˆ FEG = 105^{o} – 20^{o}

By subtraction

âˆ FEG = 85^{o}

According to the â–³ EFG

We can write

x^{o} + 25^{o} + âˆ FEG = 180^{o}

By substituting the values

x^{o} + 25^{o} + 85^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 25^{o} – 85^{o}

By subtraction

x^{o} = 180^{o} â€“ 110^{o}

x^{o} = 70^{o}

Therefore, the value of x is 70.

**12. In the given figure, AB || CD. Find the value of x.**

**Solution:**

It is given that AB || CD and AC is a transversal.

From the figure we know that âˆ BAC and âˆ ACD are consecutive interior angles

So we get

âˆ BAC + âˆ ACD = 180^{o}

By substituting the values

75^{o} + âˆ ACD = 180^{o}

On further calculation

âˆ ACD = 180^{o} – 75^{o}

By subtraction

âˆ ACD = 105^{o}

From the figure we know that âˆ ECF and âˆ ACD are vertically opposite angles

So we get

âˆ ECF = âˆ ACD = 105^{o}

According to the â–³ CEF

We can write

âˆ ECF + âˆ CEF + âˆ EFC = 180^{o}

By substituting the values

105^{o} + x^{o} + 30^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 105^{o} – 30^{o}

By subtraction

x^{o} = 180^{o} â€“ 135^{o}

x^{o} = 45^{o}

Therefore, the value of x is 45.

**13. In the given figure, AB || CD. Find the value of x.**

**Solution:**

It is given that AB || CD and PQ is a transversal

From the figure we know that âˆ PEF and âˆ EGH are corresponding angles

So we get

âˆ PEF = âˆ EGH = 85^{o}

From the figure we also know that âˆ EGH and âˆ QGH form a linear pair of angles

So we get

âˆ EGH + âˆ QGH = 180^{o}

By substituting the values we get

85^{o} + âˆ QGH = 180^{o}

On further calculation

âˆ QGH = 180^{o} – 85^{o}

By subtraction

âˆ QGH = 95^{o}

We can also find the âˆ GHQ

âˆ GHQ + âˆ CHQ = 180^{o}

By substituting the values

âˆ GHQ + 115^{o} = 180^{o}

On further calculation

âˆ GHQ = 180^{o} – 115^{o}

By subtraction

âˆ GHQ = 65^{o}

According to the â–³ GHQ

We can write

âˆ GQH + âˆ GHQ + âˆ QGH = 180^{o}

By substituting the values

x^{o} + 65^{o} + 95^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 65^{o} – 95^{o}

By subtraction

x^{o} = 180^{o} â€“ 160^{o}

x^{o} = 20^{o}

Therefore, the value of x is 20.

**14. In the given figure, AB || CD. Find the value of x, y and z.**

**Solution:**

It is given that AB || CD and BC is a transversal

From the figure we know that âˆ ABC = âˆ BCD

So we get

x = 35

It is also given that AB || CD and AD is a transversal

From the figure we know that âˆ BAD = âˆ ADC

So we get

z = 75

According to the â–³ ABO

We can write

âˆ ABO + âˆ BAO + âˆ BOA = 180^{o}

By substituting the values

x^{o} + 75^{o} + y^{o} = 180^{o}

35^{o} + 75^{o} + y^{o} = 180^{o}

On further calculation

y^{o} = 180^{o} – 35^{o} – 75^{o}

By subtraction

y^{o} = 180^{o} â€“ 110^{o}

y^{o} = 70^{o}

Therefore, the value of x, y and z is 35, 70 and 75.

**15. In the given figure, AB || CD. Prove that âˆ BAE – âˆ ECD = âˆ AEC.**

**Solution:**

Construction a line EF which is parallel to AB and CD through the point E

We know that EF || AB and AE is a transversal

From the figure we know that âˆ BAE and âˆ AEF are supplementary

So it can be written as

âˆ BAE + âˆ AEF = 180^{o}â€¦.. (1)

We also know that EF || CD and CE is a transversal

From the figure we know that âˆ DCE and âˆ CEF are supplementary

So it can be written as

âˆ DCE + âˆ CEF = 180^{o}

According to the diagram the above equation can be written as

âˆ DCE + (âˆ AEC +âˆ AEF) = 180^{o}

From equation (1) we know that âˆ AEF can be written as 180^{o} – âˆ BAE

So we get

âˆ DCE + âˆ AEC + 180^{o} – âˆ BAE = 180^{o}

So we get

âˆ BAE – âˆ DCE = âˆ AEC

Therefore, it is proved that âˆ BAE – âˆ DCE = âˆ AEC

**16. In the given figure, AB || CD. Prove that p + q â€“ r = 180.**

**Solution:**

Draw a line KH passing through the point F which is parallel to both AB and CD

We know that KF || CD and FG is a transversal

From the figure we know that âˆ KFG and âˆ FGD are alternate angles

So we get

âˆ KFG = âˆ FGD = r^{o} â€¦â€¦. (1)

We also know that AE || KF and EF is a transversal

From the figure we know that âˆ AEF and âˆ KFE are alternate angles

So we get

âˆ AEF + âˆ KFE = 180^{o}

By substituting the values we get

p^{o} + âˆ KFE = 180^{o}

So we get

âˆ KFE = 180^{o} – p^{o} â€¦â€¦. (2)

By adding both the equations (1) and (2) we get

âˆ KFG + âˆ KFE = 180^{o} – p^{o} + r^{o}

From the figure âˆ KFG + âˆ KFE can be written as âˆ EFG

âˆ EFG = 180^{o} – p^{o} + r^{o}

We know that âˆ EFG = q^{o}

q^{o} = 180^{o} – p^{o} + r^{o}

It can be written as

p + q â€“ r = 180^{o}

Therefore, it is proved that p + q â€“ r = 180^{o}

**17. In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.**

**Solution:**

From the figure we know that âˆ PRQ = x^{o} = 60^{o} as the vertically opposite angles are equal

We know that EF || GH and RQ is a transversal

From the figure we also know that âˆ PRQ and âˆ RQS are alternate angles

So we get

âˆ PRQ = âˆ RQS

âˆ x = âˆ y = 60^{o}

We know that AB || CD and PR is a transversal

From the figure we know that âˆ PRD and âˆ APR are alternate angles

So we get

âˆ PRD = âˆ APR

It can be written as

âˆ PRQ + âˆ QRD = âˆ APR

By substituting the values we get

x + âˆ QRD = 110^{o}

60^{o} + âˆ QRD = 110^{o}

On further calculation

âˆ QRD = 110^{o} – 60^{o}

By subtraction

âˆ QRD = 50^{o}

According to the â–³ QRS

We can write

âˆ QRD + âˆ QSR + âˆ RQS = 180^{o}

By substituting the values

âˆ QRD + t^{o} + y^{o} = 180^{o}

50^{o} + t^{o} + 60^{o} = 180^{o}

On further calculation

t^{o} = 180^{o} – 50^{o} – 60^{o}

By subtraction

t^{o} = 180^{o} â€“ 110^{o}

t^{o} = 70^{o}

We know that AB || CD and GH is a transversal

From the figure we know that z^{o} and t^{o} are alternate angles

So we get

z^{o} = t^{o} = 70^{o}

Therefore, the values of x, y, z and t are 60^{o}, 60^{o}, 70^{o} and 70^{o}.

**18. In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of âˆ BEF and âˆ EFD respectively, prove that âˆ EGF = 90 ^{o}.**

**Solution:**

We know that AB || CD and t is a transversal cutting at points E and F

From the figure we know that âˆ BEF and âˆ DFE are interior angles

So we get

âˆ BEF + âˆ DFE = 180^{o}

Dividing the entire equation by 2 we get

(1/2)âˆ BEF + (1/2) âˆ DFE = 90^{o}

According to the figure the above equation can further be written as

âˆ GEF + âˆ GFE = 90^{o} â€¦â€¦. (1)

According to the â–³ GEF

We can write

âˆ GEF + âˆ GFE + âˆ EGF = 180^{o}

Based on equation (1) we get

90^{o} + âˆ EGF = 180^{o}

On further calculation

âˆ EGF = 180^{o} – 90^{o}

By subtraction

âˆ EGF = 90^{o}

Therefore, it is proved that âˆ EGF = 90^{o}

**19. In the given figure, AB || CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of âˆ AEF and âˆ EFD respectively, prove that EP || FQ.**

**Solution:**

We know that AB || CD and t is a transversal

From the figure we know that âˆ AEF and âˆ EFD are alternate angles

So we get

âˆ AEF = âˆ EFD

Dividing both the sides by 2 we get

(1/2) âˆ AEF = (1/2) âˆ EFD

So we get

âˆ PEF = âˆ EFQ

The alternate interior angles are formed only when the transversal EF cuts both FQ and EP.

Therefore, it is proved that EP || FQ.

**20. In the given figure, BA || ED and BC || EF. Show that âˆ ABC = âˆ DEF.**

**Solution:**

Extend the line DE to meet the line BC at the point Z.

We know that AB || DZ and BC is a transversal

From the figure we know that âˆ ABC and âˆ DZC are corresponding angles

So we get

âˆ ABC = âˆ DZC â€¦.. (1)

We also know that EF || BC and DZ is a transversal

From the figure we know that âˆ DZC and âˆ DEF are corresponding angles

So we get

âˆ DZC = âˆ DEF â€¦â€¦ (2)

Considering both the equation (1) and (2) we get

âˆ ABC = âˆ DEF

Therefore, it is proved that âˆ ABC = âˆ DEF

**21. In the given figure, BA || ED and BC || EF. Show that âˆ ABC + âˆ DEF = 180 ^{o}.**

**Solution:**

Extend the line ED to meet the line BC at the point Z

We know that AB || EZ and BC is a transversal

From the figure we know that âˆ ABZ and âˆ EZB are interior angles

So we get

âˆ ABZ + âˆ EZB = 180^{o}

âˆ ABZ can also be written as âˆ ABC

âˆ ABC + âˆ EZB = 180^{o} â€¦â€¦ (1)

We know that EF || BC and EZ is a transversal

From the figure we know that âˆ BZE and âˆ ZEF are alternate angles

So we get

âˆ BZE = âˆ ZEF

âˆ ZEF can also be written as âˆ DEF

âˆ BZE = âˆ DEF â€¦.. (2)

By substituting equation (1) in (2) we get

âˆ ABC + âˆ DEF = 180^{o}

Therefore, it is proved that âˆ ABC + âˆ DEF = 180^{o}

**22. In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.**

**Solution:**

Construct a line m and n from A and B intersect at P

So we get

OB âŠ¥ m and OC âŠ¥ n

So m âŠ¥ n

We can also write it as

OB âŠ¥ OC

Since APB is a right angle triangle

We know that âˆ APB = 90^{o}

So we can write it as

âˆ APB = âˆ PAB + âˆ PBA

By substituting the values

90^{o} = âˆ 2 + âˆ 3

We know that angle of incidence is equal to the angle of reflection

So we get

âˆ 1 = âˆ 2 and âˆ 4 = âˆ 3

It can be written as

âˆ 1 + âˆ 4 = âˆ 2 + âˆ 3 = 90^{o}

We can write it as

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 180^{o}

We know that âˆ 1 + âˆ 2 = âˆ CAB and âˆ 3 + âˆ 4 = âˆ ABD

âˆ CAB + âˆ ABD = 180^{o}

According to the diagram âˆ CAB and âˆ ABD are consecutive interior angles when the transversal AB cuts BD and CA.

Therefore, it is proved that CA || BD.

**23. In the figure given below, state which lines are parallel and why?**

**Solution:**

According to the figure we know that âˆ BAC and âˆ ACD are alternate angles

So we get

âˆ BAC = âˆ ACD = 110^{o}

We know that âˆ BAC and âˆ ACD are alternate angles when the transversal AC cuts CD and AB.

Therefore, AB || CD.

**24. Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.**

**Solution:**

According to the figure consider the two parallel lines as m and n

We know that p âŠ¥ m and q âŠ¥ n

So we get

âˆ 1 = âˆ 2 = 90^{o}

We know that m || n and p is a transversal

From the figure we know that âˆ 1 and âˆ 3 are corresponding angles

So we get

âˆ 1 = âˆ 3

We also know that

âˆ 2 = âˆ 3 = 90^{o}

We know that âˆ 2 and âˆ 3 are corresponding angles when the transversal n cuts p and q.

So we get p || q.

Therefore, it is shown that the two lines which are perpendicular to two parallel lines are parallel to each other.

### RS Aggarwal Solutions for Class 9 Maths Chapter 7: Lines and Angles

Chapter 7, Lines and Angles, has 3 exercises from RS Aggarwal Solutions textbook, which covers entire topics to make the students ready to face the exams with higher confidence. Major topics which are explained in brief under this Chapter are as given below:

- Basic Terms and Definitions
- Various types of Angles
- Some Angle Relations
- Some Results on Angle Relations
- Results on Parallel Lines
- Corresponding Angles Axiom

### Also access RS Aggarwal Solutions Class 9 Maths Chapter 7 – Lines and Angles Exercises

Exercise 7A Solutions 12 Questions

Exercise 7B Solutions 16 Questions

Exercise 7C Solutions 24 Questions

### Chapter brief of RS Aggarwal Solutions Class 9 Maths Chapter 7 – Lines and Angles

For a better conceptual knowledge, the students can make use of RS Aggarwal Solutions for Class 9 as a reference material to perform well in the exams. The step by step description of the solutions in a simple language is done by our faculty, based on the understanding ability of the students.

Lines and Angles have a wide range of applications in various fields like designing of roads, buildings, sports facilities and lines are mainly used to plan the line of business, which gives information about the services and products offered by the company. The students can download the PDF format of the solutions attached here in order to gain a better conceptual knowledge for the board exams.