# RS Aggarwal Solutions for Class 9 Chapter 8 -Triangles

A triangle is a shape having three angles and three sides. It is mainly divided on the basis of number of sides and angles it possesses. Different types of triangles based on the number of sides are Scalene triangles, Isosceles triangles, Equilateral triangles and the triangles based on the angles are Obtuse angled triangle, Acute angled triangle and Right angled triangle. Scoring good marks in the exams has become much simpler with the help of RS Aggarwal Solutions for Class 9. The exercise wise solutions with steps based on the evaluation in the Class 9 exams improves confidence in students to solve difficult questions. RS Aggarwal Solutions for Class 9 Chapter 8 Triangles are provided here.

## Access RS Aggarwal Solutions for Class 9 Chapter 8: Triangles

Exercise 8 PAGE: 252

1. In â–³ ABC, if âˆ B = 76o and âˆ C = 48o, find âˆ A.

Solution:

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting the values in the above equation we get

âˆ A + 76o + 48o = 180o

On further calculation

âˆ A = 180o – 76o – 48o

By subtraction we get

âˆ A = 180o – 124o

âˆ A = 56o

Therefore, the value of âˆ A is 56o.

2. The angles of a triangle are in the ratio 2: 3: 4. Find the angles.

Solution:

Let us consider the measure of the angles in a triangle as 2xo, 3xo and 4xo

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

2x + 3x + 4x = 180o

9x = 180o

By division

x = 180/9

x = 20

By substituting the values of x

2xo = 2 (20) = 40o

3xo = 3 (20) = 60o

4xo = 4 (20) = 80o

Therefore, the angles are 40o, 60o and 80o

3. In â–³ ABC, if 3 âˆ A = 4 âˆ B = 6 âˆ C, calculate âˆ A, âˆ B and âˆ C.

Solution:

Consider 3 âˆ A = 4 âˆ B = 6 âˆ C = x

So we can write it as

3 âˆ A = x

âˆ A = x/3

4 âˆ B = x

âˆ B = x/4

6 âˆ C = x

âˆ C = x/6

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting the values in the above equation we get

(x/3) + (x/4) + (x/6) = 180o

LCM of 3, 4 and 6 is 12

So we get

(4x + 3x + 2x)/12 = 180o

9x/12 = 180o

By cross multiplication

9x = 180 Ã— 12

9x = 2160

By division

x = 2160/9

x = 240

By substituting the values of x

âˆ A = x/3 = 240/3 = 80o

âˆ B = x/4 = 240/4 = 60o

âˆ C = x/6 = 240/6 = 40o

Therefore, the value of âˆ A, âˆ B and âˆ C is 80o, 60o and 40o.

4. In â–³ ABC, if âˆ A + âˆ B = 108o and âˆ B + âˆ C = 130o, find âˆ A, âˆ B and âˆ C.

Solution:

It is given that âˆ A + âˆ B = 108o â€¦â€¦ (1)

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting âˆ A + âˆ B = 108o in the above equation

108o + âˆ C = 180o

On further calculation

âˆ C = 180o – 108o

By subtraction

âˆ C = 72o

It is given that âˆ B + âˆ C = 130o

By substituting the value of âˆ C

âˆ B + 72o = 130o

On further calculation

âˆ B = 130o – 72o

By subtraction

âˆ B = 58o

By substituting âˆ B = 58o in equation (1)

So we get

âˆ A + âˆ B = 108o

âˆ A + 58o = 108o

On further calculation

âˆ A = 108o – 58o

By subtraction

âˆ A = 50o

Therefore, âˆ A = 50o, âˆ B = 58o and âˆ C = 72o

5. In â–³ ABC, if âˆ A + âˆ B = 125o and âˆ A + âˆ C = 113o, find âˆ A, âˆ B and âˆ C.

Solution:

It is given that âˆ A + âˆ B = 125o â€¦. (1)

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting âˆ A + âˆ B = 125o in the above equation

125o + âˆ C = 180o

On further calculation

âˆ C = 180o – 125o

By subtraction

âˆ C = 55o

It is given that âˆ A + âˆ C = 113o

By substituting the value of âˆ C

âˆ A + 55o = 113o

On further calculation

âˆ A = 113o – 55o

By subtraction

âˆ A = 58o

By substituting âˆ A = 58o in equation (1)

So we get

âˆ A + âˆ B = 125o

58o + âˆ B= 125o

On further calculation

âˆ B = 125o – 58o

By subtraction

âˆ B = 67o

Therefore, âˆ A = 58o, âˆ B = 67o and âˆ C = 55o

6. In â–³ PQR, if âˆ P – âˆ Q = 42o and âˆ Q – âˆ R = 21o, find âˆ P, âˆ Q and âˆ R.

Solution:

It is given that âˆ P – âˆ Q = 42o

It can be written as

âˆ P = 42o + âˆ Q

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ P + âˆ Q + âˆ R = 180o

By substituting âˆ P = 42o + âˆ Q in the above equation

42o + âˆ Q +âˆ Q + âˆ R = 180o

On further calculation

42o + 2 âˆ Q + âˆ R = 180o

2 âˆ Q + âˆ R = 180o – 42o

By subtraction we get

2 âˆ Q + âˆ R = 138o â€¦. (i)

It is given that âˆ Q – âˆ R = 21o

It can be written as

âˆ R = âˆ Q – 21o

By substituting the value of âˆ R in equation (i)

2 âˆ Q + âˆ Q – 21o = 138o

On further calculation

3 âˆ Q – 21o = 138o

3 âˆ Q = 138o + 21o

3 âˆ Q = 159o

By division

âˆ Q = 159/3

âˆ Q = 53o

By substituting âˆ Q = 53o in âˆ P = 42o + âˆ Q

So we get

âˆ P = 42o + 53o

âˆ P = 95o

By substituting âˆ Q in âˆ Q – âˆ R = 21o

53o – âˆ R = 21o

On further calculation

âˆ R = 53o – 21o

By subtraction

âˆ R = 32o

Therefore, âˆ P = 95o, âˆ Q = 53o and âˆ R= 32o

7. The sum of two angles of a triangle is 116o and their difference is 24o. Find the measure of each angle of the triangle.

Solution:

Let us consider the sum of two angles as âˆ A + âˆ B = 116o and the difference can be written as âˆ A – âˆ B = 24o

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting âˆ A + âˆ B = 116o in the above equation

116o + âˆ C = 180o

On further calculation

âˆ C = 180o – 116o

By subtraction

âˆ C = 64o

It is given that âˆ A – âˆ B = 24o

It can be written as âˆ A = 24o + âˆ B

Now by substituting âˆ A = 24o + âˆ B in âˆ A + âˆ B = 116o

âˆ A + âˆ B = 116o

24o + âˆ B + âˆ B = 116o

On further calculation

24o + 2âˆ B = 116o

By subtraction

2âˆ B = 116o – 24o

2âˆ B = 92o

By division

âˆ B = 92/2

âˆ B = 46o

By substituting âˆ B = 46o in âˆ A = 24o + âˆ B

We get

âˆ A = 24o + 46o

âˆ A = 70o

Therefore, âˆ A = 70o, âˆ B = 46o and âˆ C = 64o

8. Two angles of a triangle are equal and the third angle is greater than each one of them by 18o. Find the angles.

Solution:

Consider âˆ A and âˆ B in a triangle is xo

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting the values

xo + xo + âˆ C = 180o

2xo + âˆ C = 180o â€¦.. (1)

According to the question we get

âˆ C = xo + 18o â€¦â€¦. (2)

By substituting (2) in (1) we get

2xo + xo + 18o = 180o

On further calculation

3xo + 18o = 180o

By subtraction

3xo = 180o – 18o

3xo = 162o

By division

xo = 162/3

xo = 54o

By substituting the values of x

âˆ A = âˆ B = 54o

âˆ C = 54o + 18o = 72o

Therefore, âˆ A = 54o, âˆ B = 54o and âˆ C = 72o

9. Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Solution:

Consider âˆ C is the smallest angle among âˆ ABC

According to the question

We can write it as

âˆ A = 2 âˆ C and âˆ B = 3 âˆ C

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting the values

2 âˆ C + 3 âˆ C + âˆ C = 180o

6âˆ C = 180o

By division

âˆ C = 180/6

âˆ C = 30o

Now by substituting the value of âˆ C we get

âˆ A = 2 âˆ C = 2 (30o) = 60o

âˆ B = 3 âˆ C = 3 (30o) = 90o

Therefore, âˆ A = 60o, âˆ B = 90o and âˆ C = 30o.

10. In a right-angled triangle, one of the acute angles measures 53o. Find the measure of each angle of the triangle.

Solution:

Consider ABC as a right-angled triangle with âˆ C = 90o

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

We can write it as

âˆ A + âˆ B = 180o – âˆ C

By substituting the values

âˆ A + âˆ B = 180o – 90o

By subtraction

âˆ A + âˆ B = 90o

Let us consider âˆ A = 53o

By substituting the value of âˆ A in âˆ A + âˆ B = 90o

We get

53o + âˆ B = 90o

On further calculation

âˆ B = 90o – 53o

By subtraction

âˆ B = 37o

Therefore, âˆ A = 53o, âˆ B = 37o and âˆ C = 90o.

11. If one angle of a triangle is equal to the sum of the other two, show that the triangle is right angled.

Solution:

Consider ABC as a triangle

According to the question it can be written as

âˆ A = âˆ B + âˆ C â€¦.. (1)

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

By substituting âˆ A in the above equation

âˆ B + âˆ C + âˆ B + âˆ C = 180o

2 (âˆ B + âˆ C) = 180o

By division

âˆ B + âˆ C = 180/2

âˆ B + âˆ C = 90o

According to equation (1) we can write it as

âˆ A = 90o

Therefore, it is proved that the triangle is right angled.

12. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Solution:

Consider ABC as a triangle

According to the question it can be written as

âˆ A < âˆ B + âˆ C

Add âˆ A to both the sides of the equation

So we get

âˆ A + âˆ A < âˆ A + âˆ B + âˆ C

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

So we get

2 âˆ A < 180o

By division we get

âˆ A < 180/2

âˆ A < 90o

In the same way we can also write

âˆ B < âˆ A + âˆ C

Add âˆ B to both the sides of the equation

So we get

âˆ B + âˆ B < âˆ A + âˆ B + âˆ C

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

So we get

2 âˆ B < 180o

By division we get

âˆ B < 180/2

âˆ B < 90o

So we know that

âˆ C < âˆ A + âˆ B

Add âˆ C to both the sides of the equation

So we get

âˆ C + âˆ C < âˆ A + âˆ B + âˆ C

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

So we get

2 âˆ C < 180o

By division we get

âˆ C < 180/2

âˆ C < 90o

Therefore, it is proved that the triangle ABC is acute angled.

13. If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.

Solution:

Consider ABC as a triangle

According to the question it can be written as

âˆ B > âˆ A + âˆ C â€¦. (1)

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ A + âˆ B + âˆ C = 180o

So we get

âˆ A + âˆ C = 180o – âˆ B

Substituting âˆ A + âˆ C in equation (1) we get

âˆ B > 180o – âˆ B

Add âˆ B to both the sides of the equation

So we get

âˆ B + âˆ B > 180o – âˆ B + âˆ B

2 âˆ B > 180o

By division we get

âˆ B > 180/2

âˆ B > 90o

So we know that âˆ B > 90o which means that âˆ B is an obtuse angle

Therefore, it is proved that the triangle ABC is obtuse angled.

14. In the given figure, side BC of â–³ ABC is produced to D. If âˆ ACD = 128o, and âˆ ABC = 43o, find âˆ BAC and âˆ ACB.

Solution:

From the figure we know that âˆ ACB and âˆ ACD form a linear pair of angles

So we get

âˆ ACB + âˆ ACD = 180o

By substituting the values

âˆ ACB + 128o = 180o

On further calculation

âˆ ACB = 180o – 128o

By subtraction

âˆ ACB = 52o

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ ABC + âˆ ACB + âˆ BAC = 180o

By substituting the values

43o + 52o + âˆ BAC = 180o

On further calculation

âˆ BAC = 180o – 43o – 52o

By subtraction

âˆ BAC = 180o – 95o

âˆ BAC = 85o

Therefore, âˆ BAC = 85o and âˆ ACB = 52o.

15. In the given figure, the side BC of â–³ ABC has been produced on the left-hand side from B to D and on the right-hand side from C to E. If âˆ ABD = 106o and âˆ ACE = 118o, find the measure of each angle of the triangle.

Solution:

From the figure we know that âˆ DBA and âˆ ABC form a linear pair of angles

So we get

âˆ DBA + âˆ ABC = 180o

By substituting the values

106o + âˆ ABC = 180o

On further calculation

âˆ ABC = 180o – 106o

By subtraction

âˆ ABC = 74o

From the figure we know that âˆ ACB and âˆ ACE form a linear pair of angles

So we get

âˆ ACB + âˆ ACE = 180o

By substituting the values

âˆ ACB + 118o = 180o

On further calculation

âˆ ACB = 180o – 118o

By subtraction

âˆ ACB = 62o

We know that the sum of all the angles in a triangle is 180o.

So we can write it as

âˆ ABC + âˆ ACB + âˆ BAC = 180o

By substituting the values

74o + 62o + âˆ BAC = 180o

On further calculation

âˆ BAC = 180o – 74o – 62o

By subtraction

âˆ BAC = 180o – 136o

âˆ BAC = 44o

Therefore, the measure of each angle of the triangle is âˆ A = 44o, âˆ B = 74o and âˆ C = 62o.

16. Calculate the value of x in each of the following figures.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution:

(i) From the figure we know that âˆ EAB and âˆ BAC form a linear pair of angles

So we get

âˆ EAB + âˆ BAC = 180o

By substituting the values

110o + âˆ BAC = 180o

On further calculation

âˆ BAC = 180o – 110o

By subtraction

âˆ BAC = 70o

From the figure we know that âˆ BCA and âˆ ACD form a linear pair of angles

So we get

âˆ BCA + âˆ ACD = 180o

By substituting the values

âˆ BCA + 120o = 180o

On further calculation

âˆ BAC = 180o – 120o

By subtraction

âˆ BAC = 60o

We know that the sum of all the angles in a triangle is 180o.

âˆ ABC + âˆ ACB + âˆ BAC = 180o

By substituting the values

xo + 70o + 60o = 180o

On further calculation

xo = 180o – 70o – 60o

By subtraction

xo = 180o – 130o

xo = 50o

(ii) We know that the sum of all the angles in triangle ABC is 180o.

âˆ A + âˆ B + âˆ C = 180o

By substituting the values

30o + 40o + âˆ C = 180o

On further calculation

âˆ C = 180o – 30o – 40o

By subtraction

âˆ C = 180o – 70o

âˆ C = 110o

From the figure we know that âˆ BCA and âˆ ACD form a linear pair of angles

So we get

âˆ BCA + âˆ ACD = 180o

By substituting the values

110o + âˆ ACD = 180o

On further calculation

âˆ ACD = 180o – 110o

By subtraction

âˆ ACD = 70o

We know that the sum of all the angles in triangle ECD is 180o.

âˆ ECD + âˆ CDE + âˆ CED = 180o

By substituting the values

70o + 50o + âˆ CED = 180o

On further calculation

âˆ CED = 180o – 70o – 50o

By subtraction

âˆ CED = 180o – 120o

âˆ CED = 60o

From the figure we know that âˆ AED and âˆ CED form a linear pair of angles

So we get

âˆ AED + âˆ CED = 180o

By substituting the values

xo + 60o = 180o

On further calculation

xo = 180o – 60o

By subtraction

xo = 120o

(iii) From the figure we know that âˆ EAF and âˆ BAC are vertically opposite angles

So we get

âˆ EAF = âˆ BAC = 60o

We know that in the triangle ABC exterior angle is equal to the sum of two opposite interior angles

So we can write it as

âˆ ACD = âˆ BAC + âˆ ABC

By substituting the values

115o = 60o + xo

On further calculation

xo = 115o – 60o

By subtraction

xo = 55o

(iv) We know that AB || CD and AD is a transversal

We know that the sum of all the angles in triangle ECD is 180o.

âˆ E + âˆ C + âˆ D = 180o

By substituting the values

xo + 45o + 60o = 180o

On further calculation

xo = 180o – 45o – 60o

By subtraction

xo = 180o – 105o

xo = 75o

(v) We know that in the triangle AEF exterior angle is equal to the sum of two opposite interior angles

So we can write it as

âˆ BED = âˆ EAF + âˆ EFA

By substituting the values

100o = 40o + âˆ EFA

On further calculation

âˆ EFA = 100o – 40o

By subtraction

âˆ EFA = 60o

From the figure we know that âˆ CFD and âˆ EFA are vertically opposite angles

So we get

âˆ CFD = âˆ EFA = 60o

We know that in the triangle FCD exterior angle is equal to the sum of two opposite interior angles

So we can write it as

âˆ BCD = âˆ CFD + âˆ CDF

By substituting the values

90o = 60o + xo

On further calculation

xo = 90o – 60o

By subtraction

xo = 30o

(vi) We know that the sum of all the angles in triangle ABE is 180o.

âˆ A + âˆ B + âˆ E = 180o

By substituting the values in the above equation

75o + 65o + âˆ E = 180o

On further calculation

âˆ E = 180o – 75o – 65o

By subtraction

âˆ E = 180o – 140o

âˆ E = 40o

From the figure we know that âˆ CED and âˆ AEB are vertically opposite angles

So we get

âˆ CED = âˆ AEB = 40o

We know that the sum of all the angles in triangle CED is 180o.

âˆ C + âˆ E + âˆ D = 180o

By substituting the values

110o + 40o + xo = 180o

On further calculation

xo = 180o – 110o + 40o

By subtraction

xo = 180o – 150o

xo = 30o

17. In the figure given alongside, AB || CD, EF || BC, âˆ BAC = 60o and âˆ DHF = 50o. Find âˆ GCH and âˆ AGH.

Solution:

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

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

So we get

âˆ BAC = âˆ ACD = 60o

So we also get

âˆ BAC = âˆ GCH = 60o

From the figure we also know that âˆ DHF and âˆ CHG are vertically opposite angles

So we get

âˆ DHF = âˆ CHG = 50o

We know that the sum of all the angles in triangle GCH is 180o.

So we can write it as

âˆ GCH + âˆ CHG + âˆ CGH = 180o

By substituting the values

60o + 50o + âˆ CGH = 180o

On further calculation

âˆ CGH = 180o – 60o – 50o

By subtraction

âˆ CGH = 180o – 110o

âˆ CGH = 70o

From the figure we know that âˆ CGH and âˆ AGH form a linear pair of angles

So we get

âˆ CGH + âˆ AGH = 180o

By substituting the values

70o + âˆ AGH = 180o

On further calculation

âˆ AGH = 180o – 70o

By subtraction

âˆ AGH = 110o

Therefore, âˆ GCH = 60o and âˆ AGH = 110o.

18. Calculate the value of x in the given figure.

Solution:

Construct a line CD to cut the line AB at point E.

We know that in the triangle BDE exterior angle is equal to the sum of two opposite interior angles

So we can write it as

âˆ CDB = âˆ CEB + âˆ DBE

By substituting the values

xo = âˆ CEB + 45o â€¦. (1)

We know that in the triangle AEC exterior angle is equal to the sum of two opposite interior angles

So we can write it as

âˆ CEB = âˆ CAB + âˆ ACE

By substituting the values

âˆ CEB = 55o + 30o

âˆ CEB = 85o

By substituting âˆ CEB in equation (1) we get

xo = 85o + 45o

xo = 130o

19. In the given figure, AD divides âˆ BAC in the ratio 1: 3 and AD = DB. Determine the value of x.

Solution:

It is given that AD divides âˆ BAC in the ratio 1: 3

So let us consider âˆ BAD and âˆ DAC as y and 3y

According to the figure we know that BAE is a straight line

From the figure we know that âˆ BAC and âˆ CAE form a linear pair of angles

So we get

âˆ BAC + âˆ CAE = 180o

We know that

âˆ BAC = âˆ BAD + âˆ DAC

So it can be written as

âˆ BAD + âˆ DAC + âˆ CAE = 180o

By substituting the values we get

y + 3y + 108o = 180o

On further calculation

4y = 180o – 108o

By subtraction

4y = 72o

By division

y = 72/4

y = 18o

We know that the sum of all the angles in triangle ABC is 180o.

So we can write it as

âˆ ABC + âˆ BCA + âˆ BAC = 180o

It is given that AD = DB so we can write it as âˆ ABC = âˆ BAD

From the figure we know that âˆ BAC = y + 3y = 4y

By substituting the values

y + x + 4y = 180o

On further calculation

5y + x = 180o

By substituting the value of y

5 (18o) + x = 180o

By multiplication

90o + x = 180o

x = 180o – 90o

By subtraction we get

x = 90o

Therefore, the value of x is 90.

20. If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Solution:

Given: â–³ ABC in which AB, BC and CA are produced to points D, E and F.

To prove: âˆ DCA + âˆ FAE + âˆ FBD = 180o

Proof:

From the figure we know that

âˆ DCA = âˆ A + âˆ B â€¦. (1)

âˆ FAE = âˆ B + âˆ C â€¦. (2)

âˆ FBD = âˆ A + âˆ C â€¦. (3)

By adding equation (1), (2) and (3) we get

âˆ DCA + âˆ FAE + âˆ FBD = âˆ A + âˆ B + âˆ B + âˆ C + âˆ A + âˆ C

So we get

âˆ DCA + âˆ FAE + âˆ FBD = 2 âˆ A + 2 âˆ B + 2 âˆ C

Now by taking out 2 as common

âˆ DCA + âˆ FAE + âˆ FBD = 2 (âˆ A + âˆ B + âˆ C)

We know that the sum of all the angles in a triangle is 180o.

So we get

âˆ DCA + âˆ FAE + âˆ FBD = 2 (180o)

âˆ DCA + âˆ FAE + âˆ FBD = 360o

Therefore, it is proved.

21. In the adjoining figure, show that âˆ A + âˆ B +âˆ C + âˆ D + âˆ E + âˆ F = 360o.

Solution:

We know that the sum of all the angles in triangle ACE is 180o.

âˆ A + âˆ C + âˆ E = 180o â€¦.. (1)

We know that the sum of all the angles in triangle BDF is 180o.

âˆ B + âˆ D + âˆ F = 180o â€¦.. (2)

Now by adding both equations (1) and (2) we get

âˆ A + âˆ C + âˆ E + âˆ B + âˆ D + âˆ F = 180o + 180o

On further calculation

âˆ A + âˆ B + âˆ C + âˆ D + âˆ E + âˆ F = 360o

Therefore, it is proved that âˆ A + âˆ B + âˆ C + âˆ D + âˆ E + âˆ F = 360o.

22. In the given figure, AM âŠ¥ BC and AN is the bisector of âˆ A. If âˆ ABC = 70o and âˆ ACB = 20o, find âˆ MAN.

Solution:

We know that the sum of all the angles in triangle ABC is 180o.

âˆ A + âˆ B + âˆ C = 180o

By substituting the values

âˆ A + 70o + 20o = 180o

On further calculation

âˆ A = 180o – 70o – 20o

By subtraction

âˆ A = 180o – 90o

âˆ A = 90o

We know that the sum of all the angles in triangle ABM is 180o.

âˆ BAM + âˆ ABM + âˆ AMB = 180o

By substituting the values

âˆ BAM + 70o + 90o = 180o

On further calculation

âˆ BAM = 180o – 70o – 90o

By subtraction

âˆ BAM = 180o – 160o

âˆ BAM = 20o

It is given that AN is the bisector of âˆ A

So it can be written as

âˆ BAN = (1/2) âˆ A

By substituting the values

âˆ BAN = (1/2) (90o)

By division

âˆ BAN = 45o

From the figure we know that

âˆ MAN + âˆ BAM = âˆ BAN

By substituting the values we get

âˆ MAN + 20o = 45o

On further calculation

âˆ MAN = 45o – 20o

By subtraction

âˆ MAN = 25o

Therefore, âˆ MAN = 25o.

23. In the given figure, BAD || EF, âˆ AEF = 55o and âˆ ACB = 25o, find âˆ ABC.

Solution:

We know that BAD || EF and EC is the transversal

From the figure we know that âˆ AEF and âˆ CAD are corresponding angles

So we get

âˆ AEF = âˆ CAD = 55o

From the figure we know that âˆ CAD and âˆ CAB form a linear pair of angles

So we get

âˆ CAD + âˆ CAB = 180o

By substituting the values

55o + âˆ CAB = 180o

On further calculation

âˆ CAB = 180o – 55o

By subtraction

âˆ CAB = 125o

We know that the sum of all the angles in triangle ABC is 180o.

âˆ ABC + âˆ CAB + âˆ ACB = 180o

By substituting the values in the above equation we get

âˆ ABC + 125o + 25o = 180o

On further calculation

âˆ ABC = 180o – 125o – 25o

By subtraction

âˆ ABC = 180o – 150o

âˆ ABC = 30o

24. In a â–³ ABC, it is given that âˆ A: âˆ B: âˆ C = 3: 2: 1 and CD âŠ¥ AC. Find âˆ ECD.

Solution:

In a â–³ ABC, it is given that

âˆ A: âˆ B: âˆ C = 3: 2: 1

It can also be written as

âˆ A = 3x, âˆ B = 2x and âˆ C = x

We know that the sum of all the angles in triangle ABC is 180o.

âˆ A + âˆ B + âˆ C = 180o

By substituting the values we get

3x + 2x + x = 180o

6x = 180o

By division

x = 180/6

x = 30o

Now by substituting the value of x we get

âˆ A = 3x = 3 (30o) = 90o

âˆ B = 2x = 2 (30o) = 60o

âˆ C = x = 30o

We know that in the triangle ABC exterior angle is equal to the sum of two opposite interior angles

So we can write it as

âˆ ACE = âˆ A + âˆ B

By substituting the values we get

âˆ ACE = 90o + 60o

âˆ ACE = 150o

We know that âˆ ACE can be written as âˆ ACD + âˆ ECD

So we can write it as

âˆ ACE = âˆ ACD + âˆ ECD

By substituting the values we get

150o = 90o + âˆ ECD

It is given that CD âŠ¥ AC so âˆ ACD = 90o

On further calculation

âˆ ECD = 150o – 90o

By subtraction

âˆ ECD = 60o

Therefore, âˆ ECD = 60o.

25. In the given figure, AB || DE and BD || FG such that âˆ ABC = 50o and âˆ FGH = 120o. Find the values of x and y.

Solution:

From the figure we know that âˆ FGH and âˆ FGE form a linear pair of angles

So we get

âˆ FGH + âˆ FGE = 180o

By substituting the values

120o + y = 180o

On further calculation

y = 180o – 120o

By subtraction

y = 60o

We know that AB || DF and BD is a transversal

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

So we get

âˆ ABC = âˆ CDE = 50o

We know that BD || FG and DF is the transversal

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

So we get

âˆ EFG = âˆ CDE = 50o

We know that the sum of all the angles in triangle EFG is 180o.

âˆ FEG + âˆ FGE + âˆ EFG = 180o

By substituting the values we get

x + y + 50o = 180o

x + 60o + 50o = 180o

On further calculation

x = 180o – 60o – 50o

By subtraction

x = 180o – 110o

x = 70o

Therefore, the values of x = 70o and y = 60o.

26. In the given figure, AB || CD and EF is a transversal. If âˆ AEF = 65o, âˆ DFG = 30o, âˆ EGF = 90o and âˆ GEF = xo, find the value of x.

Solution:

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

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

So we get

âˆ AEF = âˆ EFG + âˆ DFG

By substituting the values

65o = âˆ EFG + 30o

On further calculation

âˆ EFG = 65o – 30o

By subtraction

âˆ EFG = 35o

We know that the sum of all the angles in triangle GEF is 180o.

âˆ GEF + âˆ EGF + âˆ EFG = 180o

By substituting the values we get

x + 90o + 35o = 180o

On further calculation

x = 180o – 90o – 35o

By subtraction

x = 55o

Therefore, the value of x is 55o

27. In the given figure, AB || CD, âˆ BAE = 65o and âˆ OEC = 20o. Find âˆ ECO.

Solution:

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

From the figure we know that âˆ BAE and âˆ DOE are corresponding angles

So we get

âˆ BAE = âˆ DOE = 65o

From the figure we know that âˆ DOE and âˆ COE form a linear pair of angles

So we get

âˆ DOE + âˆ COE = 180o

By substituting the values

65o + âˆ COE = 180o

On further calculation

âˆ COE = 180o – 65o

By subtraction

âˆ COE = 115o

We know that the sum of all the angles in triangle OCE is 180o.

âˆ OEC + âˆ ECO + âˆ COE = 180o

By substituting the values we get

20o + âˆ ECO + 115o = 180o

On further calculation

âˆ ECO = 180o – 20o – 115o

By subtraction

âˆ ECO = 45o

Therefore, âˆ ECO = 45o

28. In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If âˆ EGB = 35o and QP âŠ¥ EF, find the measure of âˆ PQH.

Solution:

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

From the figure we know that âˆ EGB and âˆ GHD are corresponding angles

So we get

âˆ EGB = âˆ GHD = 35o

From the figure we know that âˆ GHD and âˆ QHP are vertically opposite angles

So we get

âˆ GHD = âˆ QHP = 35o

We know that the sum of all the angles in triangle DQHP is 180o.

âˆ PQH + âˆ QHP + âˆ QPH = 180o

By substituting the values we get

âˆ PQH + 35o + 90o = 180o

On further calculation

âˆ PQH = 180o – 35o – 90o

By subtraction

âˆ PQH = 180o – 125o

âˆ PQH = 55o

Therefore, âˆ PQH = 55o

29. In the given figure, AB || CD and EF âŠ¥ AB. If EG is the transversal such that âˆ GED = 130o, find âˆ EGF.

Solution:

We know that AB || CD and GE is the transversal

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

So we get

âˆ EGF + âˆ GED = 180o

By substituting the values

âˆ EGF + 130o = 180o

On further calculation

âˆ EGF = 180o – 130o

By subtraction

âˆ EGF = 50o

Therefore, âˆ EGF = 50o

### RS Aggarwal Solutions for Class 9 Maths Chapter 8: Triangles

Chapter 8, Triangles, has 1 exercise with solutions prepared by our expert team of faculties having much knowledge about the concepts in Mathematics. Some of the topics in RS Aggarwal Solutions which are explained in brief under this chapter are:

• Triangles
• Types of triangles on the basis of sides
• Types of triangles on the basis of angles
• Some terms related to triangle
• Exterior and interior opposite angles of a triangle
• Some results on triangles

### RS Aggarwal Solutions Class 9 Maths Chapter 8 – Triangles

Students can make use of the RS Aggarwal Solutions which covers the entire syllabus as per the textbook with exercise wise answers in each chapter. The brief solutions prepared by our experts mainly help students to solve the problems of higher difficulty with ease. RS Aggarwal Solutions for Class 9 can be used by the students as a vital resource to boost their exam preparation. Some of the applications of triangles are architecture, design of buildings, height of bridges and finding the area of geometric structures. Here, we provide PDF containing chapter wise solutions of the entire chapter based on the RS Aggarwal textbook to help students study well before appearing for the exams.