RD Chapter 11- Co-ordinate Geometry Ex-11.2 |
RD Chapter 11- Co-ordinate Geometry Ex-VSAQS |
RD Chapter 11- Co-ordinate Geometry Ex-MCQS |

**Answer
1** :

∵ Sum of three angles of a triangle is 180°

∴ In ∆ABC, ∠A = 55°, ∠B = 40°

But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ 55° + 40° + ∠C = 180°

⇒ 95° + ∠C = 180°

∴ ∠C= 180° -95° = 85°

**Answer
2** :

Ratio in three angles of a triangle =1:2:3

Let first angle = x

Then second angle = 2x

and third angle = 3x

∴ x + 2x + 3x = 180° (Sum of angles of a triangle)

⇒6x = 180°

⇒x == 30°

∴ First angle = x = 30°

Second angle = 2x = 2 x 30° = 60°

and third angle = 3x = 3 x 30° = 90°

∴ Angles are 30°, 60°, 90°

**Answer
3** :

∵ Sum of three angles of a triangle = 180°

∴ (x –40)° + (x – 20)° + (x-10)0 = 180°

⇒ x –40° + x – 20° + x – 10° = 180°

⇒ x +x+ x – 70° = 180°

⇒ x = 180° + 70° = 250°

⇒ x= = 100°

∴ x =100°

**Answer
4** :

Let each of the two equal angles = x

Then third angle = x + 30°

But sum of the three angles of a triangle is 180°

∴ x + x + x + 30° = 180°

⇒ 3x + 30° = 180°

⇒3x = 150° ⇒x = = 50°

∴ Each equal angle = 50°

and third angle = 50° + 30° = 80°

∴ Angles are 50°, 50° and 80°

**Answer
5** :

In the triangle ABC,

∠B = ∠A + ∠C

But ∠A + ∠B + ∠C = 180°

⇒∠B + ∠A + ∠C = 180°

⇒∠B + ∠B = 180°

⇒2∠B = 180°

∴ ∠B = = 90°

∵ One angle of the triangle is 90°

∴ ∆ABC is a right triangle.

Can a triangle have:

(i) Two right angles?

(ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60°?

(v) All angles less than 60°?

(vi) All angles equal to 60°?

Justify your answer in each case.

**Answer
6** :

(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180° and if there are two right-angles, then the third angle will be zero which is not possible.

(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180° and if there are

two obtuse angle, then the third angle will be negative which is not possible.

(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.

(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.

(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.

(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.

**Answer
7** :

Let three angles of a triangle be x°, (x + 10)°, (x + 20)°

But sum of three angles of a triangle is 180°

∴ x + (x+ 10)° + (x + 20) = 180°

⇒ x + x+10°+ x + 20 = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° – 30° = 150°

∴ x = = 50°

∴ Angle are 50°, 50 + 10, 50 + 20

i.e. 50°, 60°, 70°

**Answer
8** :

In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O

Now ∠B + ∠C = 180° – 12° = 108°

∵ OB and OC are the bisectors of ∠B and ∠C respectively

∴ ∠OBC + ∠OCB = (B + C)

= x 108° = 54°

But in ∆OBC,

∴ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 54° + ∠BOC = 180°

∠BOC = 180°-54°= 126°

OR

According to corollary,

∠BOC = 90°+ ∠A

= 90+ x 72° = 90° + 36° = 126°

**Answer
9** :

In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠C respectively

To prove : ∠BOC cannot be a right angle

Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively

∴ ∠BOC = 90° x ∠A

Let ∠BOC = 90°, then

∠A = O

⇒∠A = O

Which is not possible because the points A, B and C will be on the same line Hence, ∠BOC cannot be a right angle.

**Answer
10** :

Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°

To prove : ∆ABC is a right angled triangle

Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O

∴ ∠BOC = 90°+ ∠A

But ∠BOC =135°

∴ 90°+ ∠A = 135°

⇒ ∠A= 135° -90° = 45°

∴ ∠A = 45° x 2 = 90°

∴ ∆ABC is a right angled triangle

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