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.

## RS Aggarwal Solutions for Class 9 Chapter 8: Triangles Download PDF

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

Exercise 8 PAGE: 252

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

**Solution:**

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

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values in the above equation we get

âˆ A + 76^{o} + 48^{o} = 180^{o}

On further calculation

âˆ A = 180^{o} – 76^{o} – 48^{o}

By subtraction we get

âˆ A = 180^{o} – 124^{o}

âˆ A = 56^{o}

Therefore, the value of âˆ A is 56^{o}.

**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 2x^{o}, 3x^{o} and 4x^{o}

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

So we can write it as

2x + 3x + 4x = 180^{o}

By addition

9x = 180^{o}

By division

x = 180/9

x = 20

By substituting the values of x

2x^{o} = 2 (20) = 40^{o}

3x^{o} = 3 (20) = 60^{o}

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

Therefore, the angles are 40^{o}, 60^{o} and 80^{o}

**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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values in the above equation we get

(x/3) + (x/4) + (x/6) = 180^{o}

LCM of 3, 4 and 6 is 12

So we get

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

By addition

9x/12 = 180^{o}

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 = 80^{o}

âˆ B = x/4 = 240/4 = 60^{o}

âˆ C = x/6 = 240/6 = 40^{o}

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

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

**Solution:**

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

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

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

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

108^{o} + âˆ C = 180^{o}

On further calculation

âˆ C = 180^{o} – 108^{o}

By subtraction

âˆ C = 72^{o}

It is given that âˆ B + âˆ C = 130^{o}

By substituting the value of âˆ C

âˆ B + 72^{o} = 130^{o}

On further calculation

âˆ B = 130^{o} – 72^{o}

By subtraction

âˆ B = 58^{o}

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

So we get

âˆ A + âˆ B = 108^{o}

âˆ A + 58^{o} = 108^{o}

On further calculation

âˆ A = 108^{o }– 58^{o}

By subtraction

âˆ A = 50^{o}

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

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

**Solution:**

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

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

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

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

125^{o} + âˆ C = 180^{o}

On further calculation

âˆ C = 180^{o} – 125^{o}

By subtraction

âˆ C = 55^{o}

It is given that âˆ A + âˆ C = 113^{o}

By substituting the value of âˆ C

âˆ A + 55^{o} = 113^{o}

On further calculation

âˆ A = 113^{o} – 55^{o}

By subtraction

âˆ A = 58^{o}

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

So we get

âˆ A + âˆ B = 125^{o}

58^{o} + âˆ B= 125^{o}

On further calculation

âˆ B = 125^{o }– 58^{o}

By subtraction

âˆ B = 67^{o}

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

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

**Solution:**

It is given that âˆ P – âˆ Q = 42^{o}

It can be written as

âˆ P = 42^{o }+ âˆ Q

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

So we can write it as

âˆ P + âˆ Q + âˆ R = 180^{o}

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

42^{o} + âˆ Q +âˆ Q + âˆ R = 180^{o}

On further calculation

42^{o} + 2 âˆ Q + âˆ R = 180^{o}

2 âˆ Q + âˆ R = 180^{o} – 42^{o}

By subtraction we get

2 âˆ Q + âˆ R = 138^{o} â€¦. (i)

It is given that âˆ Q – âˆ R = 21^{o}

It can be written as

âˆ R = âˆ Q – 21^{o}

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

2 âˆ Q + âˆ Q – 21^{o} = 138^{o}

On further calculation

3 âˆ Q – 21^{o} = 138^{o}

3 âˆ Q = 138^{o }+ 21^{o}

By addition

3 âˆ Q = 159^{o}

By division

âˆ Q = 159/3

âˆ Q = 53^{o}

By substituting âˆ Q = 53^{o} in âˆ P = 42^{o }+ âˆ Q

So we get

âˆ P = 42^{o }+ 53^{o}

By addition

âˆ P = 95^{o}

By substituting âˆ Q in âˆ Q – âˆ R = 21^{o}

53^{o} – âˆ R = 21^{o}

On further calculation

âˆ R = 53^{o} – 21^{o}

By subtraction

âˆ R = 32^{o}

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

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

**Solution:**

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

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

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

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

116^{o} + âˆ C = 180^{o}

On further calculation

âˆ C = 180^{o} – 116^{o}

By subtraction

âˆ C = 64^{o}

It is given that âˆ A – âˆ B = 24^{o}

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

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

âˆ A + âˆ B = 116^{o}

24^{o} + âˆ B + âˆ B = 116^{o}

On further calculation

24^{o} + 2âˆ B = 116^{o}

By subtraction

2âˆ B = 116^{o} – 24^{o}

2âˆ B = 92^{o}

By division

âˆ B = 92/2

âˆ B = 46^{o}

By substituting âˆ B = 46^{o} in âˆ A = 24^{o} + âˆ B

We get

âˆ A = 24^{o} + 46^{o}

By addition

âˆ A = 70^{o}

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

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

**Solution:**

Consider âˆ A and âˆ B in a triangle is x^{o}

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

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values

x^{o} + x^{o} + âˆ C = 180^{o}

By addition

2x^{o} + âˆ C = 180^{o} â€¦.. (1)

According to the question we get

âˆ C = x^{o} + 18^{o} â€¦â€¦. (2)

By substituting (2) in (1) we get

2x^{o} + x^{o} + 18^{o} = 180^{o}

On further calculation

3x^{o} + 18^{o} = 180^{o}

By subtraction

3x^{o }= 180^{o} – 18^{o}

3x^{o }= 162^{o}

By division

x^{o} = 162/3

x^{o} = 54^{o}

By substituting the values of x

âˆ A = âˆ B = 54^{o}

âˆ C = 54^{o} + 18^{o} = 72^{o}

Therefore, âˆ A = 54^{o}, âˆ B = 54^{o} and âˆ C = 72^{o}

**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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values

2 âˆ C + 3 âˆ C + âˆ C = 180^{o}

By addition

6âˆ C = 180^{o}

By division

âˆ C = 180/6

âˆ C = 30^{o}

Now by substituting the value of âˆ C we get

âˆ A = 2 âˆ C = 2 (30^{o}) = 60^{o}

âˆ B = 3 âˆ C = 3 (30^{o}) = 90^{o}

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

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

**Solution:**

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

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

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

We can write it as

âˆ A + âˆ B = 180^{o} – âˆ C

By substituting the values

âˆ A + âˆ B = 180^{o} – 90^{o}

By subtraction

âˆ A + âˆ B = 90^{o}

Let us consider âˆ A = 53^{o}

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

We get

53^{o} + âˆ B = 90^{o}

On further calculation

âˆ B = 90^{o} – 53^{o}

By subtraction

âˆ B = 37^{o}

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

**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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting âˆ A in the above equation

âˆ B + âˆ C + âˆ B + âˆ C = 180^{o}

2 (âˆ B + âˆ C) = 180^{o}

By division

âˆ B + âˆ C = 180/2

âˆ B + âˆ C = 90^{o}

According to equation (1) we can write it as

âˆ A = 90^{o}

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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

So we get

2 âˆ A < 180^{o}

By division we get

âˆ A < 180/2

âˆ A < 90^{o}

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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

So we get

2 âˆ B < 180^{o}

By division we get

âˆ B < 180/2

âˆ B < 90^{o}

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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

So we get

2 âˆ C < 180^{o}

By division we get

âˆ C < 180/2

âˆ C < 90^{o}

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 180^{o}.

So we can write it as

âˆ A + âˆ B + âˆ C = 180^{o}

So we get

âˆ A + âˆ C = 180^{o} – âˆ B

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

âˆ B > 180^{o} – âˆ B

Add âˆ B to both the sides of the equation

So we get

âˆ B + âˆ B > 180^{o} – âˆ B + âˆ B

By addition we get

2 âˆ B > 180^{o}

By division we get

âˆ B > 180/2

âˆ B > 90^{o}

So we know that âˆ B > 90^{o} 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 = 128 ^{o}, and âˆ ABC = 43^{o}, 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 = 180^{o}

By substituting the values

âˆ ACB + 128^{o} = 180^{o}

On further calculation

âˆ ACB = 180^{o} – 128^{o}

By subtraction

âˆ ACB = 52^{o}

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

So we can write it as

âˆ ABC + âˆ ACB + âˆ BAC = 180^{o}

By substituting the values

43^{o} + 52^{o} + âˆ BAC = 180^{o}

On further calculation

âˆ BAC = 180^{o} – 43^{o} – 52^{o}

By subtraction

âˆ BAC = 180^{o} – 95^{o}

âˆ BAC = 85^{o}

Therefore, âˆ BAC = 85^{o} and âˆ ACB = 52^{o}.

**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 = 106 ^{o} and âˆ ACE = 118^{o}, 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 = 180^{o}

By substituting the values

106^{o} + âˆ ABC = 180^{o}

On further calculation

âˆ ABC = 180^{o} – 106^{o}

By subtraction

âˆ ABC = 74^{o}

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

So we get

âˆ ACB + âˆ ACE = 180^{o}

By substituting the values

âˆ ACB + 118^{o} = 180^{o}

On further calculation

âˆ ACB = 180^{o} – 118^{o}

By subtraction

âˆ ACB = 62^{o}

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

So we can write it as

âˆ ABC + âˆ ACB + âˆ BAC = 180^{o}

By substituting the values

74^{o} + 62^{o} + âˆ BAC = 180^{o}

On further calculation

âˆ BAC = 180^{o} – 74^{o} – 62^{o}

By subtraction

âˆ BAC = 180^{o} – 136^{o}

âˆ BAC = 44^{o}

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

**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 = 180^{o}

By substituting the values

110^{o} + âˆ BAC = 180^{o}

On further calculation

âˆ BAC = 180^{o} – 110^{o}

By subtraction

âˆ BAC = 70^{o}

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

So we get

âˆ BCA + âˆ ACD = 180^{o}

By substituting the values

âˆ BCA + 120^{o} = 180^{o}

On further calculation

âˆ BAC = 180^{o} – 120^{o}

By subtraction

âˆ BAC = 60^{o}

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

âˆ ABC + âˆ ACB + âˆ BAC = 180^{o}

By substituting the values

x^{o} + 70^{o} + 60^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 70^{o} – 60^{o}

By subtraction

x^{o} = 180^{o} – 130^{o}

x^{o} = 50^{o}

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

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values

30^{o} + 40^{o} + âˆ C = 180^{o}

On further calculation

âˆ C = 180^{o} – 30^{o} – 40^{o}

By subtraction

âˆ C = 180^{o} – 70^{o}

âˆ C = 110^{o}

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

So we get

âˆ BCA + âˆ ACD = 180^{o}

By substituting the values

110^{o} + âˆ ACD = 180^{o}

On further calculation

âˆ ACD = 180^{o} – 110^{o}

By subtraction

âˆ ACD = 70^{o}

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

âˆ ECD + âˆ CDE + âˆ CED = 180^{o}

By substituting the values

70^{o} + 50^{o }+ âˆ CED = 180^{o}

On further calculation

âˆ CED = 180^{o} – 70^{o} – 50^{o}

By subtraction

âˆ CED = 180^{o} – 120^{o}

âˆ CED = 60^{o}

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

So we get

âˆ AED + âˆ CED = 180^{o}

By substituting the values

x^{o} + 60^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 60^{o}

By subtraction

x^{o} = 120^{o}

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

So we get

âˆ EAF = âˆ BAC = 60^{o}

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

115^{o} = 60^{o} + x^{o}

On further calculation

x^{o} = 115^{o} – 60^{o}

By subtraction

x^{o} = 55^{o}

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

So we get âˆ BAD = âˆ ADC = 60^{o}

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

âˆ E + âˆ C + âˆ D = 180^{o}

By substituting the values

x^{o} + 45^{o} + 60^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 45^{o} – 60^{o}

By subtraction

x^{o} = 180^{o} – 105^{o}

x^{o} = 75^{o}

(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

100^{o} = 40^{o} + âˆ EFA

On further calculation

âˆ EFA = 100^{o} – 40^{o}

By subtraction

âˆ EFA = 60^{o}

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

So we get

âˆ CFD = âˆ EFA = 60^{o}

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

90^{o} = 60^{o} + x^{o}

On further calculation

x^{o} = 90^{o} – 60^{o}

By subtraction

x^{o} = 30^{o}

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

âˆ A + âˆ B + âˆ E = 180^{o}

By substituting the values in the above equation

75^{o} + 65^{o} + âˆ E = 180^{o}

On further calculation

âˆ E = 180^{o} – 75^{o} – 65^{o}

By subtraction

âˆ E = 180^{o} – 140^{o}

âˆ E = 40^{o}

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

So we get

âˆ CED = âˆ AEB = 40^{o}

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

âˆ C + âˆ E + âˆ D = 180^{o}

By substituting the values

110^{o} + 40^{o} + x^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 110^{o} + 40^{o}

By subtraction

x^{o} = 180^{o} – 150^{o}

x^{o} = 30^{o}

**17. In the figure given alongside, AB || CD, EF || BC, âˆ BAC = 60 ^{o} and âˆ DHF = 50^{o}. 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 = 60^{o}

So we also get

âˆ BAC = âˆ GCH = 60^{o}

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

So we get

âˆ DHF = âˆ CHG = 50^{o}

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

So we can write it as

âˆ GCH + âˆ CHG + âˆ CGH = 180^{o}

By substituting the values

60^{o} + 50^{o} + âˆ CGH = 180^{o}

On further calculation

âˆ CGH = 180^{o} – 60^{o} – 50^{o}

By subtraction

âˆ CGH = 180^{o} – 110^{o}

âˆ CGH = 70^{o}

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

So we get

âˆ CGH + âˆ AGH = 180^{o}

By substituting the values

70^{o} + âˆ AGH = 180^{o}

On further calculation

âˆ AGH = 180^{o} – 70^{o}

By subtraction

âˆ AGH = 110^{o}

Therefore, âˆ GCH = 60^{o} and âˆ AGH = 110^{o}.

**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

x^{o} = âˆ CEB + 45^{o} 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 = 55^{o} + 30^{o}

By addition

âˆ CEB = 85^{o}

By substituting âˆ CEB in equation (1) we get

x^{o} = 85^{o} + 45^{o}

By addition

x^{o} = 130^{o}

**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 = 180^{o}

We know that

âˆ BAC = âˆ BAD + âˆ DAC

So it can be written as

âˆ BAD + âˆ DAC + âˆ CAE = 180^{o}

By substituting the values we get

y + 3y + 108^{o} = 180^{o}

On further calculation

4y = 180^{o} – 108^{o}

By subtraction

4y = 72^{o}

By division

y = 72/4

y = 18^{o}

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

So we can write it as

âˆ ABC + âˆ BCA + âˆ BAC = 180^{o}

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 = 180^{o}

On further calculation

5y + x = 180^{o}

By substituting the value of y

5 (18^{o}) + x = 180^{o}

By multiplication

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

x = 180^{o} – 90^{o}

By subtraction we get

x = 90^{o}

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 = 180^{o}

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 180^{o}.

So we get

âˆ DCA + âˆ FAE + âˆ FBD = 2 (180^{o})

âˆ DCA + âˆ FAE + âˆ FBD = 360^{o}

Therefore, it is proved.

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

**Solution:**

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

âˆ A + âˆ C + âˆ E = 180^{o} â€¦.. (1)

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

âˆ B + âˆ D + âˆ F = 180^{o} â€¦.. (2)

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

âˆ A + âˆ C + âˆ E + âˆ B + âˆ D + âˆ F = 180^{o} + 180^{o}

On further calculation

âˆ A + âˆ B + âˆ C + âˆ D + âˆ E + âˆ F = 360^{o}

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

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

**Solution:**

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

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values

âˆ A + 70^{o} + 20^{o} = 180^{o}

On further calculation

âˆ A = 180^{o} – 70^{o} – 20^{o}

By subtraction

âˆ A = 180^{o} – 90^{o}

âˆ A = 90^{o}

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

âˆ BAM + âˆ ABM + âˆ AMB = 180^{o}

By substituting the values

âˆ BAM + 70^{o} + 90^{o} = 180^{o}

On further calculation

âˆ BAM = 180^{o} – 70^{o} – 90^{o}

By subtraction

âˆ BAM = 180^{o} – 160^{o}

âˆ BAM = 20^{o}

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) (90^{o})

By division

âˆ BAN = 45^{o}

From the figure we know that

âˆ MAN + âˆ BAM = âˆ BAN

By substituting the values we get

âˆ MAN + 20^{o} = 45^{o}

On further calculation

âˆ MAN = 45^{o} – 20^{o}

By subtraction

âˆ MAN = 25^{o}

Therefore, âˆ MAN = 25^{o}.

**23. In the given figure, BAD || EF, âˆ AEF = 55 ^{o} and âˆ ACB = 25^{o}, 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 = 55^{o}

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

So we get

âˆ CAD + âˆ CAB = 180^{o}

By substituting the values

55^{o} + âˆ CAB = 180^{o}

On further calculation

âˆ CAB = 180^{o} – 55^{o}

By subtraction

âˆ CAB = 125^{o}

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

âˆ ABC + âˆ CAB + âˆ ACB = 180^{o}

By substituting the values in the above equation we get

âˆ ABC + 125^{o} + 25^{o} = 180^{o}

On further calculation

âˆ ABC = 180^{o} – 125^{o} – 25^{o}

By subtraction

âˆ ABC = 180^{o} – 150^{o}

âˆ ABC = 30^{o}

**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 180^{o}.

âˆ A + âˆ B + âˆ C = 180^{o}

By substituting the values we get

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

By addition

6x = 180^{o}

By division

x = 180/6

x = 30^{o}

Now by substituting the value of x we get

âˆ A = 3x = 3 (30^{o}) = 90^{o}

âˆ B = 2x = 2 (30^{o}) = 60^{o}

âˆ C = x = 30^{o}

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 = 90^{o} + 60^{o}

By addition

âˆ ACE = 150^{o}

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

150^{o} = 90^{o} + âˆ ECD

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

On further calculation

âˆ ECD = 150^{o} – 90^{o}

By subtraction

âˆ ECD = 60^{o}

Therefore, âˆ ECD = 60^{o}.

**25. In the given figure, AB || DE and BD || FG such that âˆ ABC = 50 ^{o} and âˆ FGH = 120^{o}. 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 = 180^{o}

By substituting the values

120^{o} + y = 180^{o}

On further calculation

y = 180^{o} – 120^{o}

By subtraction

y = 60^{o}

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 = 50^{o}

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 = 50^{o}

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

âˆ FEG + âˆ FGE + âˆ EFG = 180^{o}

By substituting the values we get

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

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

On further calculation

x = 180^{o} – 60^{o} – 50^{o}

By subtraction

x = 180^{o} – 110^{o}

x = 70^{o}

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

**26. In the given figure, AB || CD and EF is a transversal. If âˆ AEF = 65 ^{o}, âˆ DFG = 30^{o}, âˆ EGF = 90^{o} and âˆ GEF = x^{o}, 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

65^{o} = âˆ EFG + 30^{o}

On further calculation

âˆ EFG = 65^{o} – 30^{o}

By subtraction

âˆ EFG = 35^{o}

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

âˆ GEF + âˆ EGF + âˆ EFG = 180^{o}

By substituting the values we get

x + 90^{o} + 35^{o} = 180^{o}

On further calculation

x = 180^{o} – 90^{o} – 35^{o}

By subtraction

x = 55^{o}

Therefore, the value of x is 55^{o}

**27. In the given figure, AB || CD, âˆ BAE = 65 ^{o} and âˆ OEC = 20^{o}. 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 = 65^{o}

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

So we get

âˆ DOE + âˆ COE = 180^{o}

By substituting the values

65^{o} + âˆ COE = 180^{o}

On further calculation

âˆ COE = 180^{o} – 65^{o}

By subtraction

âˆ COE = 115^{o}

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

âˆ OEC + âˆ ECO + âˆ COE = 180^{o}

By substituting the values we get

20^{o} + âˆ ECO + 115^{o} = 180^{o}

On further calculation

âˆ ECO = 180^{o} – 20^{o} – 115^{o}

By subtraction

âˆ ECO = 45^{o}

Therefore, âˆ ECO = 45^{o}

**28. In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If âˆ EGB = 35 ^{o} 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 = 35^{o}

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

So we get

âˆ GHD = âˆ QHP = 35^{o}

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

âˆ PQH + âˆ QHP + âˆ QPH = 180^{o}

By substituting the values we get

âˆ PQH + 35^{o} + 90^{o} = 180^{o}

On further calculation

âˆ PQH = 180^{o} – 35^{o} – 90^{o}

By subtraction

âˆ PQH = 180^{o} – 125^{o}

âˆ PQH = 55^{o}

Therefore, âˆ PQH = 55^{o}

**29. In the given figure, AB || CD and EF âŠ¥ AB. If EG is the transversal such that âˆ GED = 130 ^{o}, 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 = 180^{o}

By substituting the values

âˆ EGF + 130^{o} = 180^{o}

On further calculation

âˆ EGF = 180^{o} – 130^{o}

By subtraction

âˆ EGF = 50^{o}

Therefore, âˆ EGF = 50^{o}

### 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 – Exercise list

Exercise 8 Solutions 29 Questions

### 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.