In this chapter, we provide RD Sharma Solutions for Chapter 11 Coordinate Geometry Ex 11.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 11 Coordinate Geometry Ex 11.2 Maths pdf, free RD Sharma Solutions for Chapter 11 Coordinate Geometry Ex 11.2 Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 11 |

Chapter Name | Coordinate Geometry |

Exercise | Ex 11.2 |

**RD Sharma Solutions for Class 9 Chapter 11 Coordinate Geometry Ex 11.2 Download PDF**

Question 1.

The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

Solution:

In ∆ABC, base BC is produced both ways to D and E respectivley forming ∠ABE = 104° and ∠ACD = 136°

Question 2.

In the figure, the sides BC, CA and AB of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ∆ABC.

Solution:

In ∆ABC, sides BC, CA and BA are produced to D, E and F respectively.

∠ACD = 105° and ∠EAF = 45°

∠ACD + ∠ACB = 180° (Linear pair)

⇒ 105° + ∠ACB = 180°

⇒ ∠ACB = 180°- 105° = 75°

∠BAC = ∠EAF (Vertically opposite angles)

= 45°

But ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 45° + ∠ABC + 75° = 180°

⇒ 120° +∠ABC = 180°

⇒ ∠ABC = 180°- 120°

∴ ∠ABC = 60°

Hence ∠ABC = 60°, ∠BCA = 75°

and ∠BAC = 45°

Question 3.

Compute the value of x in each of the following figures:

Solution:

(i) In ∆ABC, sides BC and CA are produced to D and E respectively

(ii) In ∆ABC, side BC is produced to either side to D and E respectively

∠ABE = 120° and ∠ACD =110°

∵ ∠ABE + ∠ABC = 180° (Linear pair)

(iii) In the figure, BA || DC

Question 4.

In the figure, AC ⊥ CE and ∠A: ∠B : ∠C = 3:2:1, find the value of ∠ECD.

Solution:

In ∆ABC, ∠A : ∠B : ∠C = 3 : 2 : 1

BC is produced to D and CE ⊥ AC

∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangles)

Let∠A = 3x, then ∠B = 2x and ∠C = x

∴ 3x + 2x + x = 180° ⇒ 6x = 180°

⇒ x = 180∘6 = 30°

∴ ∠A = 3x = 3 x 30° = 90°

∠B = 2x = 2 x 30° = 60°

∠C = x = 30°

In ∆ABC,

Ext. ∠ACD = ∠A + ∠B

⇒ 90° + ∠ECD = 90° + 60° = 150°

∴ ∠ECD = 150°-90° = 60°

Question 5.

In the figure, AB || DE, find ∠ACD.

Solution:

In the figure, AB || DE

AE and BD intersect each other at C ∠BAC = 30° and ∠CDE = 40°

∵ AB || DE

∴ ∠ABC = ∠CDE (Alternate angles)

⇒ ∠ABC = 40°

In ∆ABC, BC is produced

Ext. ∠ACD = Int. ∠A + ∠B

= 30° + 40° = 70°

Question 6.

Which of the following statements are true (T) and which are false (F):

(i) Sum of the three angles of a triangle is 180°.

(ii) A triangle can have two right angles.

(iii) All the angles of a triangle can be less than 60°.

(iv) All the angles of a triangle can be greater than 60°.

(v) All the angles of a triangle can be equal to 60°.

(vi) A triangle can have two obtuse angles.

(vii) A triangle can have at most one obtuse angles.

(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.

(ix) An exterior angle of a triangle is less than either of its interior opposite angles.

(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

(xi) An exterior angle of a triangle is greater than the opposite interior angles.

Solution:

(i) True.

(ii) False. A right triangle has only one right angle.

(iii) False. In this, the sum of three angles will be less than 180° which is not true.

(iv) False. In this, the sum of three angles will be more than 180° which is not true.

(v) True. As sum of three angles will be 180° which is true.

(vi) False. A triangle has only one obtuse angle.

(vii) True.

(viii)True.

(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.

(x) True.

(xi) True.

Question 7.

Fill in the blanks to make the following statements true:

(i) Sum of the angles of a triangle is ………

(ii) An exterior angle of a triangle is equal to the two …….. opposite angles.

(iii) An exterior angle of a triangle is always …….. than either of the interior opposite angles.

(iv) A triangle cannot have more than ………. right angles.

(v) A triangles cannot have more than ……… obtuse angles.

Solution:

(i) Sum of the angles of a triangle is 180°.

(ii) An exterior angle of a triangle is equal to the two interior opposite angles.

(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.

(iv) A triangle cannot have more than one right angles.

(v) A triangles cannot have more than one obtuse angles.

Question 8.

In a ∆ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.

Solution:

Given : In ∆ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior ∠B and ∠C meet at P and bisectors of exterior angles B and C meet at Q.

To prove : ∠BPC + ∠BQC = 180°

Proof : ∵ PB and PC are the internal bisectors of ∠B and ∠C

∠BPC = 90°+ 12 ∠A …(i)

Similarly, QB and QC are the bisectors of exterior angles B and C

∴ ∠BQC = 90° + 12 ∠A …(ii)

Adding (i) and (ii),

∠BPC + ∠BQC = 90° + 12 ∠A + 90° – 12 ∠A

= 90° + 90° = 180°

Hence ∠BPC + ∠BQC = 180°

Question 9.

In the figure, compute the value of x.

Solution:

In the figure,

∠ABC = 45°, ∠BAD = 35° and ∠BCD = 50° Join BD and produce it E

Question 10.

In the figure, AB divides ∠D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Solution:

In the figure AB = DB

Question 11.

ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = 12 ∠A.

Solution:

Given : In ∠ABC, CB is produced to E bisectors of ext. ∠ABE and into ∠ACB meet at D.

Question 12.

In the figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Solution:

Question 13.

In a AABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.

Solution:

Given : In ∆ABC,

∠C > ∠B and AD is the bisector of ∠A

To prove : ∠ADB > ∠ADC

Proof: In ∆ABC, AD is the bisector of ∠A

∴ ∠1 = ∠2

In ∆ADC,

Ext. ∠ADB = ∠l+ ∠C

⇒ ∠C = ∠ADB – ∠1 …(i)

Similarly, in ∆ABD,

Ext. ∠ADC = ∠2 + ∠B

⇒ ∠B = ∠ADC – ∠2 …(ii)

From (i) and (ii)

∵ ∠C > ∠B (Given)

∴ (∠ADB – ∠1) > (∠ADC – ∠2)

But ∠1 = ∠2

∴ ∠ADB > ∠ADC

Question 14.

In ∆ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180°-∠A.

Solution:

Given : In ∆ABC, BD ⊥ AC and CE⊥ AB BD and CE intersect each other at O

To prove : ∠BOC = 180° – ∠A

Proof: In quadrilateral ADOE

∠A + ∠D + ∠DOE + ∠E = 360° (Sum of angles of quadrilateral)

⇒ ∠A + 90° + ∠DOE + 90° = 360°

∠A + ∠DOE = 360° – 90° – 90° = 180°

But ∠BOC = ∠DOE (Vertically opposite angles)

⇒ ∠A + ∠BOC = 180°

∴ ∠BOC = 180° – ∠A

Question 15.

In the figure, AE bisects ∠CAD and ∠B = ∠C. Prove that AE || BC.

Solution:

Given : In AABC, BA is produced and AE is the bisector of ∠CAD

∠B = ∠C

To prove : AE || BC

Proof: In ∆ABC, BA is produced

∴ Ext. ∠CAD = ∠B + ∠C

⇒ 2∠EAC = ∠C + ∠C (∵ AE is the bisector of ∠CAE) (∵ ∠B = ∠C)

⇒ 2∠EAC = 2∠C

⇒ ∠EAC = ∠C

But there are alternate angles

∴ AE || BC

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