# Class 9 RD Sharma Solutions – Chapter 9 Triangles and its Angles- Exercise 9.1

### Question 1: In a triangle Î”ABC, if âˆ A = 55Â° âˆ B = 40Â°, find âˆ C.

Solution:

Given: A = 55Â° and B = 40Â°

Theorem Used: Sum of all angles of a triangle is 180Â°.

From the theorem we can write that:

âˆ A + âˆ B + âˆ C = 180Â°

55Â° + 40Â° + âˆ C = 180Â° //Putting the values of A and B.

95Â° + âˆ C = 180Â°

âˆ C = 180Â° – 95Â°

âˆ C = 85Â°

The angle âˆ C is 85Â°.

### Question 2: If the angles of a triangle are in the ratio 1:2:3, determine three angles.

Solution:

Given: Angles of a triangle are in the ratio 1:2:3

Let the angles be x, 2x, 3x

Theorem Used: Sum of all angles of a triangle is 180Â°.

x + 2x + 3x = 180Â°

6x = 180Â°

x = 180Â°/6

x = 30Â°   //Deriving the value of x

Deriving the value of the other two angles from the value of x

2x = 2 X (30Â°) = 60Â°

3x = 3 X (30Â°) = 90Â°

All the three angles are 30Â°,60Â° and 90Â° respectively.

### Question 3: The angles of a triangle are (x âˆ’ 40)Â°, (x âˆ’ 20)Â° and (1/2 x âˆ’ 10)Â°. Find the value of x.

Solution:

The angles of a triangle are (x âˆ’ 40)Â°, (x âˆ’ 20)Â° and (1/2x âˆ’ 10)Â°

Theorem Used: Sum of all angles of a triangle is 180Â°.

(x âˆ’ 40)Â° + (x âˆ’ 20) Â° + (1/2 x âˆ’ 10)Â° = 180Â°

5/2 x â€“ 70Â° = 180Â°

5/2 x = 180Â° + 70Â°

5x = 2(250)Â°

x = 500Â°/5

x = 100Â°

The value of x is 100Â°

### Question 4: The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10Â°, find the three angles.

Solution:

Given: The difference between two consecutive angles is 10Â°.

Theorem Used: Sum of all angles of a triangle is 180Â°.

Let the smallest angle of the triangle is xÂ°.

Hence, according to the given condition, the other two consecutive angles are (x + 10)Â° and (x + 20)Â° respectively.

Now from the theorem mentioned we can write that:

x + (x + 10Â°) + (x + 20Â°) = 180Â°.

3x + 30Â° = 180Â° //Simplifying the equation

3x = 180Â° – 30Â°

3x = 150Â°

x = 150Â°/3

x = 50Â°

Hence, here we get the smallest angle is 50Â°.

The next consecutive angles are, 50Â° + 10Â° = 60Â° and 50Â° + 20Â° = 70Â° respectively.

Hence, the three angles of the triangle are 50Â°, 60Â°, and 70Â° respectively.

### Question 5: Two angles of a triangle are equal and the third angle is greater than each of those angles by 30Â°. Determine all the angles of the triangle.

Solution:

Given: (i) Two angles of a triangle are equal

(ii)The third angle is greater than each of those angles by 30Â°

Theorem Used: Sum of all angles of a triangle is 180Â°.

Let the equal angles are xÂ° and the other angle is (x+30)Â°.

Now from the theorem mentioned we can write that:

x + x + (x + 30Â°) = 180Â°

3x + 30Â° = 180Â°

3x = 180Â° – 30Â°

3x = 150Â°

x = 150Â°/3

x = 50Â°

Hence, the equal angles are 50Â° each and the other angle is (50 + 30)Â° = 80Â°.

The angles of the triangle are 50Â°, 50Â° and 80Â° respectively.

### Question 6: If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right-angle triangle.

Solution:

Given: one angle of a triangle is equal to the sum of the other two

Theorem Used: Sum of all angles of a triangle is 180Â°.

Let the three angles of the triangle are âˆ A, âˆ B and âˆ (A+B) respectively.

âˆ A + âˆ B + âˆ (A + B) = 180Â°

2(âˆ A + âˆ B) = 180Â°

âˆ A + âˆ B = 90Â°  //Hence the third angle A + B = 90Â° (Proved)

### Question 7: ABC is a triangle in which angle âˆ A = 72Â°. The internal bisector of angle âˆ B and âˆ C meet in O. Find the magnitude of âˆ BOC.

Solution:

Given: (i) âˆ A = 72Â° of the triangle ABC

(ii)Internal bisectors of angle âˆ B and âˆ C meet in point O.

Theorems Used: Sum of all three angles of a triangle is 180Â°

In triangle ABC,

âˆ A + âˆ B + âˆ C = 180Â°

72Â° + âˆ B + âˆ C = 180Â°

âˆ B + âˆ C = 180Â° – 72Â° = 108Â°

âˆ B/2 + âˆ C/2 = 108Â°/2 = 54Â°                         // Dividing both sides by 2

âˆ OBC + âˆ OCB = 54Â°  –derivation (1)           // From the triangle we can clearly see this as OB and OC are the angle bisectors

Now in â–³BOC,

âˆ OBC + âˆ OCB + âˆ BOC = 180Â°

âˆ BOC + (âˆ OBC + âˆ OCB) = 180Â°

âˆ BOC + 54Â° = 180Â°    //Putting the value of âˆ OBC + âˆ OCB = 54Â° from derivation(1)

âˆ BOC = 180Â° – 54Â° = 126Â° (Ans)

âˆ BOC = 126Â°.

### Question 8: The bisectors of the base angles of a triangle cannot enclose a right angle at any case.

Solution:

Theorems Used: Sum of all three angles of a triangle is 180Â°

From the triangle â–³ABC,

âˆ A + âˆ B + âˆ C = 180Â°

âˆ A/2 + âˆ B /2 + âˆ C/2 = 180Â°/2 = 90Â°   //Dividing both sides by 2

âˆ B /2 + âˆ C/2 = 90Â° – âˆ A/2  —-(1)

From the triangle â–³BOC,

âˆ BOC + âˆ OBC + âˆ OCB = 180Â°

As OB and OC are the angle bisectors, âˆ OBC = âˆ B/2 and âˆ OCB = âˆ C/2.

âˆ BOC + âˆ B /2+ âˆ C/2 = 180Â°          //Putting the values of âˆ B /2+ âˆ C/2  = 90Â° – âˆ A/2     from the previous derivation,

âˆ BOC + 90Â° – âˆ A/2 = 180Â°

âˆ BOC = 180Â° – 90Â° + âˆ A/2 = 90Â° + âˆ A/2

For any valid triangle â–³ABC âˆ A > 0,  It implies that âˆ A/2 > 0,

That simply means

âˆ BOC not equals to 90Â° at any case. (proved)

### Question 9: If the bisectors of base angles of a triangle enclose an angle of 135Â°, Prove that the triangle is a right-angle triangle.

Solution:

Given: In â–³BOC the âˆ BOC = 135Â°

Theorems Used: Sum of all three angles of a triangle is 180Â°

From the triangle â–³ABC,

âˆ A + âˆ B + âˆ C = 180Â°

âˆ A/2 + âˆ B /2+ âˆ C/2 = 180Â°/2 = 90Â°   //Dividing both sides by 2

âˆ B /2+ âˆ C/2 = 90Â° – âˆ A/2  —-(1)

From the triangle â–³BOC,

âˆ BOC + âˆ OBC + âˆ OCB = 180Â°

As OB and OC are the angle bisectors, âˆ OBC = âˆ B/2 and âˆ OCB = âˆ C/2.

âˆ BOC + âˆ B/2 + âˆ C/2 = 180Â°          //Putting the values of âˆ B/2 + âˆ C/2  = 90Â° – âˆ A/2     from the previous derivation,

âˆ BOC +  90Â° – âˆ A/2 = 180Â°

âˆ BOC = 180Â° – 90Â° + âˆ A/2 = 90Â° + âˆ A/2

Putting the value âˆ BOC = 135Â° from the given condition,

90Â° + âˆ A/2 = 135Â°

âˆ A/2 = 135Â° – 90Â° = 45Â°

âˆ A = 45Â° X 2 = 90Â°

Hence, â–³ABC is a right angle triangle(proved)

### Question 10: In a triangle â–³ABC, âˆ ABC = âˆ ACB and the bisector of âˆ ABC and âˆ ACB intersect at O such that âˆ BOC = 120Â°. Show that âˆ A = âˆ B = âˆ C = 60Â°.

Solution:

Given: (i)âˆ ABC = âˆ ACB

(ii)âˆ BOC = 120Â°

From the triangle â–³ABC,

âˆ ABC = âˆ ACB

âˆ ABC/2 = âˆ ACB/2

âˆ OBC = âˆ OCB

From the triangle â–³ABC,

âˆ OBC + âˆ OCB + âˆ BOC = 180Â°

From the given condition âˆ BOC = 120Â°, and âˆ OBC = âˆ OCB

We can write that,

âˆ OBC + âˆ OBC + 120Â° = 180Â°.

2 X âˆ OBC = 180Â° – 120Â° = 60Â°

âˆ ABC = 60Â°

As angle âˆ ACB = âˆ ABC,

âˆ ACB = 60Â°

âˆ ACB + âˆ ABC + âˆ BAC = 180Â°

60Â° + 60Â° + âˆ BAC = 180Â°

âˆ BAC = 180Â° – 120Â° = 60Â°

Hence,

âˆ A = âˆ B = âˆ C = 60Â°.(Proved)

### (i) Two right angles.

If the triangle have two right angles, sum of those angles become 90Â° + 90Â° = 180Â° , That implies that the size of the third angle is 180Â° – 180Â° = 0, That is not possible,

### (ii)Two obtuse angles

The size of an obtuse angle is greater than 90Â° , Hence the sum of both angles are greater than 180Â° , But we know sum of all three angles of a triangle is 180Â°. So it is not possible.

### (iii) Two acute angles

Having two acute angles does not violates any law as their  sum is less than 180Â°

### (iv) All Angles More than 60Â°

Having all angles more than 60Â° will make the sum of all angles > 180Â°.But we know sum of all three angles of a triangle is 180Â° . So it is not possible.

### (v) All Angles Less than 60Â°

Having all angles more than 60Â° will make the sum of all angles < 180Â°. But we know sum of all three angles of a triangle is 180Â° . So it is not possible.

### (vi) All angles equal to 60Â°

Having all angles equal to 60Â° will make the sum of all angles = 180Â°.And we know sum of all three angles of a triangle is 180Â° . So it is possible.

### Question 12: If each angle of a triangle is less than the sum of the other two, Show that all angles of the triangle are acute angle.

Solution:

Given: Each angle is less than the sum of the other two

âˆ A + âˆ B + âˆ C = 180Â°

Given that ,âˆ A < âˆ B + âˆ C, So we can write,

âˆ A < 90Â°,

The same thing can be done for âˆ B and âˆ C.

Hence, it is proved that all the three angles are acute angle.

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