RD Chapter 11- Co-ordinate Geometry Ex-11.1 |
RD Chapter 11- Co-ordinate Geometry Ex-11.2 |
RD Chapter 11- Co-ordinate Geometry Ex-VSAQS |

If all the three angles of a triangle are equal, then each one of them is equal to

(a) 90°

(b) 45°

(c) 60°

(d) 30°

**Answer
1** :

∵ Sum of three angles of a triangle = 180°

∴ Eachangle = = 60° (c)

If two acute angles of a right triangle are equal, then each acute is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 90°

**Answer
2** :

In a right triangle, one angle = 90°

∴ Sum of other two acute angles = 180° – 90° = 90°

∵ Both angles are equal

∴ Each angle will be = = 45° (b)

An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to

(a) 75°

(b) 80°

(c) 40°

(d) 50°

**Answer
3** :

In atriangle, exterior angles is equal to the sum of its interior opposite angles

∴ Sum ofinterior opposite angles = 100°

∵ Bothangles are equal

∴ Eachangle will be = = 50° (d)

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

(a) an isosceles triangle

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangle

**Answer
4** :

Let ∠A, ∠B, ∠C be the angles of a ∆ABC and let ∠A = ∠B + ∠C

But ∠A + ∠B + ∠C = 180°

( Sum of angles of a triangle)

∴ ∠A + ∠A = 180° ⇒ 2∠A = 180°

⇒ ∠A = = 90°

∴ ∆ is a right triangle (d)

Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = 12∠A, then ∠A is equal to

(a) 80°

(b) 75°

(c) 60°

(d) 90°

**Answer
5** :

Side BC of ∆ABC is produced to D, then

Ext. ∠ACB = ∠A + ∠B

(Exterior angle of a triangle is equal to the sum of its interior opposite angles)

In ∆ABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =

(a) 35°

(b) 90°

(c) 70°

(d) 55°

**Answer
6** :

In ∆ABC, ∠B = ∠C

AX is the bisector of ext. ∠CAD

∠DAX = 70°

∴ ∠DAC = 70° x 2 = 140°

But Ext. ∠DAC = ∠B + ∠C

= ∠C + ∠C (∵ ∠B = ∠C)

= 2∠C

∴ 2∠C = 140° ⇒ ∠C = = 70°

∴ ∠ACB = 70° (c)

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is

(a) 55°

(b) 85°

(c) 40°

(d) 9.0°

**Answer
7** :

In ∆ABC, BA is produced to D such that ∠CAD = 95°

and let ∠C = 55° and ∠B = x°

∵ Exterior angle of a triangle is equal to the sum of its opposite interior angle

∴ ∠CAD = ∠B + ∠C ⇒ 95° = x + 55°

⇒ x = 95° – 55° = 40°

∴ Other interior angle = 40° (c)

If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is

(a) 90°

(b) 180°

(c) 270°

(d) 360°

**Answer
8** :

In ∆ABC, sides AB, BC and CA are produced in order, then exterior ∠FAB, ∠DBC and ∠ACE are formed

We know an exterior angles of a triangle is equal to the sum of its interior opposite angles

∴ ∠FAB = ∠B + ∠C

∠DBC = ∠C + ∠A and

∠ACE = ∠A + ∠B Adding we get,

∠FAB + ∠DBC + ∠ACE = ∠B + ∠C + ∠C + ∠A + ∠A + ∠B

= 2(∠A + ∠B + ∠C)

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

= 360° (d)

In ∆ABC, if ∠A = 100°, AD bisects ∠A and AD⊥ BC. Then, ∠B =

(a) 50°

(b) 90°

(c) 40°

(d) 100°

**Answer
9** :

In ∆ABC, ∠A = 100°

AD is bisector of ∠A and AD ⊥ BC

Now, ∠BAD == 50°

In ∆ABD,

∠BAD + ∠B + ∠D= 180°

(Sum of angles of a triangle)

⇒ ∠50° + ∠B + 90° = 180°

∠B + 140° = 180°

⇒ ∠B = 180° – 140° ∠B = 40° (c)

An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4:5. The angles of the triangle are

(a) 48°, 60°, 72°

(b) 50°, 60°, 70°

(c) 52°, 56°, 72°

(d) 42°, 60°, 76°

**Answer
10** :

In ∆ABC, BC is produced to D and ∠ACD = 108°

Ratio in ∠A : ∠B = 4:5

∵ Exterior angle of a triangle is equal to the sum of its opposite interior angles

∴ ∠ACD = ∠A + ∠B = 108°

Ratio in ∠A : ∠B = 4:5

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