Read RD Sharma Solutions Class 9 Chapter 9 Triangle and its Angles below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

**Exercise 9.1**

**Question 1: In a ΔABC, if ∠A = 55°, ∠B = 40°, find ∠C.**

**Solution:**

Given that, ∠A = 55° and ∠B = 40°

∠A + ∠B + ∠C = 180° (angle sum property)

Put the value of ∠A and ∠B in above equation.

55° + 40° + ∠C = 180°

95° + ∠C = 180°

∠C = 180° − 95°

∠C = 85°

Thus, the value of ∠C is 85°

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

**Solution:**

Let the angles be x,2x,3x

Angle sum property says that all angles of triangle is 180°.

x+2x+3x=180°

6x=180°

x=(180°)/6

x=30°

x=30°

2x=2(30)° =60°

3x= 3(30)° =90°

Hence, the values of angles are 30°, 60° and 90°.

**Question 3: The angles of a triangle are (x − 40)°, (x − 20)° and (1/2 x-10°). Find the value of x.**

**Solution:**

It is given that, angles of a triangle are(x-40)°, (x-20)° and (1/2 x-10)°

Angle sum property says that all angles of triangle is 180°.

(x-40)°+(x-20)°+(1/2 x-10)°= 180°

2x+1/2 x-40°-20°-10°=180°

5/2 x – 70° = 180°

5/2 x = 180° + 70°

5x = 2(250)°

x = (500°)/5

x = 100°

Hence, the value of x is 100°.

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

**Solution:**

It is given that the difference between two consecutive angles is 10°.

Let the first angle will be x,

Second angle will be x + 10°

Third angle will be x + 20°

Angle sum property says that all angles of triangle is 180°.

x + (x + 10°) + (x + 20°) = 180°

x + x + 10°+ x + 20° = 180°

3x + 30° = 180°

3x = 180°– 30°

3x = 150°

x = (150°)/3

x = 50°

x + 10° = 50° + 10° = 60°

x + 20° = 50° + 20° = 70°

Hence, the value of angles are 50°, 60° and 70°.

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

**Solution:**

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°.

Let the first angle and second angle will be x

Third angle will be x + 30°

Angle sum property says that all angles of triangle is 180°.

x + x + (x + 30°) = 180°

x + x + x + 30° = 180°

3x + 30° = 180°

3x = 150°

3x = (150°)/3

x = 50°

First and second angle x = 50°

Third angle = x + 30°

Third angle = 50° + 30°

Third angle = 80°

Hence, the three angles of a triangle are 50°, 50° and 80°.

**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:**

It is given that one angle of a triangle is equal to the sum of the other two angles.

Let x, y and z are three angles of a triangle.

z = x + y_______(1)

Angle sum property says that all angles of triangle is 180°.

x + y + z = 180°

z + z = 180°

2z = 180°

z = 90°

One angle of a triangle is 90° hence, it is right triangle.

**Question 7: ABC is a triangle in which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.**

**Solution:**

It is given that, ABC is a triangle where ∠A = 72°

In ΔABC,

Angle sum property says that all angles of triangle is 180°.

∠A + ∠B + ∠C = 180°

72° + ∠B + ∠C = 180°

∠B + ∠C = 180° − 72°

Line OB is the bisector of ∠B and Line OC is bisector of ∠C.

Dividing both sides by '2'

(∠B)/2 + (∠C)/2 = (180°)/2

∠OBC + ∠OCB = 54°

In ΔBOC

Angle sum property says that all angles of triangle is 180°.

∠OBC + ∠OCB + ∠BOC = 180°

54° + ∠BOC = 180°

∠BOC = 180° − 54°

∠BOC = 126°

Hence, the value of ∠BOC is 126°.

**Question 8: The bisectors of base angles of a triangle cannot enclose a right angle in any case.**

**Solution:**

It is given that ∆XYZ in which YO and CO are the bisectors of the base angles ∠B and ∠C respectively.

Let us assumed that ∠YOZ = 90°

∠ZYO + ∠YZO = 180°

1/2∠Y + 1/2∠Z = 1/2×180°

∠Y + ∠C = 180°

∠A = 0° (as per triangle sum property)

This shows that the points A, B and C do not form a triangle, which is false. So our assumption is wrong.

Hence, ∠YOZ is not a right triangle.

**Question: 9 If the bisectors of the base angles of a triangle enclose an angle of 135°, prove that the triangle is a right angle.**

**Solution:**

It is given that the bisectors of the base angles of a triangle enclose an angle of 135°

∠BOC = 135° (bisectors of the base angles)

But, We know that

∠BOC = 90° + 1/2∠A

135° = 90° + 1/2∠A

135° - 90° = 1/2∠A

45° = 1/2∠A

45° × 2 = ∠A

∠A = 90°

Hence, ΔABC is a right angle triangle that is right angled at A.

**Question 10: In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.**

**Solution:**

It is given that,

∠ABC = ∠ACB

Bisectors in ΔABC,

1/2∠ABC = 1/2∠ACB

OB, OC bisects ∠B and ∠C

∠OBC = ∠OCB

∠BOC = 90° + 1/2∠A

120° = 90°+ 1/2∠A

120° - 90° = 1/2∠A

30° × 2 = ∠A

∠A = 60°

In ΔABC

Sum of all angles of a triangle is 180°.

∠A + ∠ABC + ∠ACB = 180°

60° + 2∠ABC = 180° (∠ABC = ∠ACB)

2∠ABC = 180° − 60°

∠ABC = (120°)/2

∠ABC = 60°

∠ABC = ∠ACB

∠ACB = 60°

Hence it is proved that ∠ACB is 60°.

**Question 11: 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.**

**Solution:**

(i) No

(ii) No

(iii) Yes

(iv) No

(v) No

(vi) Yes

**Question 12: If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.**

**Solution:**

It is given that, each angle of a triangle less than the sum of the other two

∠A < ∠B + ∠C

∠B < ∠C + ∠A

∠C < ∠A + ∠B

∠A + ∠B + ∠C = 180° (Angle sum property)

∠A < ∠B + ∠C

Add both side ∠A

∠A +∠A < ∠A + ∠B + ∠C

2∠A < 180°

∠A < 90°

Similarly,

∠B < 90°

∠C < 90°

Hence, it is proved that the triangles are acute angle.

**Exercise 9.2**

**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:**

It is given that, DBC is a line.

∠DBA + ∠ABC = 180°

104° + ∠ABC = 180°

∠ABC = 180° - 104°

∠ABC = 76°

∠ACE + ∠ACB = 180°

135° + ∠ACB = 180°

∠ACB = 180° - 135°

∠ABC = 45°

By the angle sum property

∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + 76° + 45° = 180°

∠BAC + 121° = 180°

∠BAC = 180° - 121°

∠BAC = 59°

Hence, the value of angles of a triangle are ∠A = 59°, ∠B = 76° and ∠C = 45°.

**Question 2: In a triangle 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:**

Let us assumed that,

∠ABD = 2x and ∠ACE = 2y

∠ABC = 180° − 2x (linear pair)

∠ACB = 180° − 2y (linear pair)

∠A + ∠ABC + ∠ACB = 180° [Sum of all angles of a triangle]

∠A + 180° − 2x + 180° − 2y = 180°

−∠A + 2x + 2y = 180°

x + y = 90° + 1/2∠A

Now in ΔBQC

x + y + ∠BQC = 180° [Sum of all angles of a triangle]

90° + 1/2∠A + ∠BQC = 180°

∠BQC = 90° - 1/2∠A_____(1)

Adding (i) and (ii) we get ∠BPC + ∠BQC = 180°

Hence proved.

**Question 3: In 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:**

It is given that,

∠EAF = 45°

∠BAC = ∠EAF (Vertically opposite angles)

∠BAC = 45°

∠ACB + ∠ACD = 180° (Linear pair)

∠ACB + 105° = 180°

∠ACD = 180° – 105°

∠ACD = 75°

By the angle sum property, we get

∠BAC + ∠ACB + ∠ABC = 180°

45° + 75° + ∠ABC = 180°

105° + ∠ABC = 180°

∠ABC = 105° – 45°

∠ABC = 60°

Hence the value of ∠ABC = 60°, ∠ACD = 75° and ∠BAC = 45°.

**Question 4: Compute the value of x in each of the following figures:**

**(i)**

**Solution:**

EAC is a line,

∠EAB + ∠BAC = 180° (Linear Pair)

120° + ∠BAC = 180°

∠BAC = 180° - 120°

∠BAC = 60°

BCD is also a line

∠ACB + ∠ACD = 180° (Linear Pair)

∠ACB + 112° = 180°

∠ACB = 180° - 112°

∠ACB = 180° - 112°

∠ACB = 68°

Sum of all angles of a triangle is 180°

Let ∠ABC is x

∠ABC + ∠BAC + ∠ACB = 180°

x + 60° + 68° = 180°

x + 128° = 180°

x=180°-128°

x=52°

Hence, the value of x is 52°

**Solution:**

It is given that, EBCD lies on the same base so it is a line.

∠ABE + ∠ABC = 180° (Linear pair)

120° + ∠ABC = 180°

∠ABC = 180° - 120°

∠ABC = 60°

∠ACD + ∠ACB = 180° (Linear pair)

110° + ∠ACB = 180°

∠ACB = 180° - 110°

∠ACB = 70°

Sum of all angles of a triangle is 180°

∠ABC + ∠ACB + ∠BAC = 180°

60° + 70° + ∠BAC = 180°

130° + ∠BAC = 180°

∠BAC = 180° − 130°

∠BAC = 50°

∠BAC = x

x = 50°

Hence, the value of x is 50°

**Solution:**

∠BAE = ∠EDC = 52° (Alternate angles)

Sum of all angles of a triangle is 180°.

∠ECD + ∠EDC + ∠CED = 180°

40° + 52° + ∠CED = 180°

92° + ∠CED = 180°

∠CED = 180° - 92°

∠CED = 88°

∠CED = x

x = 88°

Hence, the value of angle x is 88°.

**Solution:**

In triangle ∆ABE,

∠BAE + ∠ABE + ∠BEA = 180° (angle sum property)

35° + 45° + ∠BEA = 180°

80° + ∠BEC = 180°

∠BEC = 180° - 80°

∠BEC = 100°

∠BED + ∠DEC = 180°

100° + ∠DEC = 180°

∠DEC = 180° - 100°

∠DEC = 80°

By exterior angle property

∠CDA = ∠DEC + ∠DCE

∠CDA = 80° + 50°

∠CDA = 130°

Hence, the value of angle ∠CDA is 130°.

**Question 5: In figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.**

**Solution:**

EAD is a line,

It is given that in figure,

∠EAC = 108°

∠DAC + ∠EAC = 180°

∠DAC + 108° = 180°

∠DAC = 180° − 108°

∠DAC = 72°

(∠BAC)/(∠DAB) = 1/3

3∠BAC = ∠DAB

∠DAC = ∠BAC + ∠DAB

72° = ∠BAC + ∠DAB

Put the value of ∠DAB as 3∠BAC.

∠BAC + 3∠BAC = 72°

4∠BAC = 72°

∠BAC = 72/4

∠BAC = 18°

So we get

∠DAB = 3 × 18°

∠DAB = 54°

∠DAB = ∠BDA = 54° (as AB = DB)

∠ABD = 180° − (54° + 54°)

∠ABD = 180 − 108°

∠ABD = 72°

Now,

∠DBA = 72° = ∠BAC + x (Exterior angle)

So we get

x =72°−18°

x = 54°

Hence, the value of angle x is 54°.

**Question 6: 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 = 1/2∠A.**

**Solution:**

Let ∠ABE = 2x and ∠ACB = 2y

ABC is a triangle ∠ABC = 180° − 2x [Linear pair]

∠A = 180° − ∠ABC − ∠ACB [Angle sum property]

= 180° − 180° + 2x + 2y

= 2(x − y) ..... (i)

Now, ∠D = 180° − ∠DBC − ∠DCB

∠D = 180° − (x + 180° − 2x) − y

∠D = 180° − x − 180° + 2x − y

(x − y)

**Question 7: In figure 9.36, AC ⊥ CE and ∠A:∠B: ∠C = 3: 2: 1, find**

**Solution:**

It is given that,

∠A : ∠B : ∠C = 3 : 2 : 1

Let us assumed that the angles be 3x, 2x and x

∠A + ∠B + ∠C = 180° (Angle sum property)

3x + 2x + x = 180°

6x = 180°

x = 30°

∠ACB = x = 30°

∠ABC = 2x = 2 × 30°

∠ABC = 60°

From the figure we can that,

∠ACE = 90°

BCD is a line

∠BCA + ∠ACE + ∠ECD = 180°

30° + 90° + ∠ECD = 180°

120° + ∠ECD = 180°

∠ECD = 180°- 120°

∠ECD = 60°

Hence, the value of ∠ ECD is 60°.

**Question 8: In figure 9.37, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.**

**Solution:**

Let us assumed that,

∠BAN = ∠NAC = x (AN bisects ∠A)

∠ANM = x + 33° (Exterior angle property)

In ΔAMB

∠BAM = 90° − 65° (Exterior angle property)

∠BAM = 25°

∠MAN = ∠BAN − ∠BAM

∠MAN = x − 25°

In ΔMAN,

∠MAN + ∠ANM + ∠AMN = 180° (Angle sum property)

(x - 25)° + (x + 33)° + 90° = 180°

x - 25° + x + 33° + 90° = 180°

2x + 8° + 90° = 180°

2x + 98° = 180°

2x = 180° - 98°

2x = 82°

x = (82°)/2

x = 41°

∠MAN = x − 25°

∠MAN = 41° − 25°

∠MAN = 16°

Hence, the value of ∠MAN is 16°.

**Question 9: In a triangle ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.**

**Solution:**

It is given that,

∠C > ∠B __________(1)

∠A + ∠C + ∠ ADC = 180° (Linear Pair)

x + ∠C + ∠ ADC = 180°

x + ∠C = 180° - ∠ ADC

∠A + ∠B + ∠ ADB = 180° (Linear Pair)

x + ∠B + ∠ ADB = 180°

x + ∠B = 180° - ∠ ADB

Adding x on both sides of equation 1.

∠C + x > ∠B + x

180° - ∠ ADC > 180° - ∠ADB

- ∠ ADC > - ∠ADB

∠ADB > ∠ADC

Hence proved

**Question 10: In triangle ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° - ∠A.**

**Solution:**

From the figure we know that, AEOD is a quadrilateral.

∠A + ∠AEO + ∠EOD + ∠ADO = 360° (Sum of all four angle of a quadrilateral is 360°)

∠A + 90° + 90° + ∠EOD = 360°

∠A + 180° + ∠EOD = 360°

∠A + ∠EOD = 360° - 180°

∠A + ∠EOD = 180°

∠EOD = ∠BOC (vertically opposite angles)

∠A + ∠BOC = 180°

∠BOC = 180° − ∠A

Hence Proved

**Question 11: In figure 9.38, AE bisects ∠CAD and ∠B = ∠C. Prove that AE ∥ BC.**

**Solution:**

Let us assumed that ∠B = ∠C = x

We know that,

∠CAD = ∠B + ∠C

∠CAD = x + x

∠CAD = 2x (exterior angle is sum of interior angles)

(∠CAD)/2 = x (we know that, ∠CAD = ∠CAE + ∠EAD)

∠EAC = x

∠EAC = ∠C

Hence it is proved that, these are the alternate interior angles for the lines AE and BC. AE ∥ BC

**Question 12: In figure 9.39, AB ∥ DE. Find ∠ACD.**

**Solution:**

It is given that, AB ∥ DE

∠ABC = ∠CDE = 40° (Alternate angles)

∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property)

40° + ∠ACB + 30° = 180°

∠ACB = 180° − 40° − 30°

∠ACB = 110°

∠ACD = 180° − 110° (Linear pair)

∠ACD = 70°

Hence, the value of ∠ACD is 70°.

**Question 13; 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) T

(ii) F

(iii) F

(iv) F

(v) T

(vi) F

(vii) T

(viii) T

(ix) F

(x) T

(xi) T

**Question 14 : 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) 180°

(ii) Interior

(iii) Greater

(iv) One

(v) One

**Exercise VSAQs ........................**

**Question 1: Define a triangle.**

**Solution:**

A triangle is a three-sided polygon with three vertices and three sides. The number of a triangle's internal angles equals 180 degrees is the most essential property of a triangle.

**Question 2: Write the sum of the angles of an obtuse triangle.**

**Solution:**

180° is the sum of angles of obtuse triangle.

**Question 3: In △ABC, if ∠B = 60°, ∠C = 800 and the bisectors of angles ∠ABC and ∠ACB meet at point O, then find the measure of ∠BOC.**

**Solution:**

It is given that,

∠B = 60°, ∠C = 80°

As per question:

∠OBC = (60°)/2

∠OBC = 30°

∠OCB = (80°)/2

∠OCB = 40°

In ∆BOC,

We know that, Sum of angles of a triangle is 180°

∠OBC + ∠OCB + ∠BOC = 180° (angle sum property)

30° + 40° + ∠BOC = 180°

70° + ∠BOC = 180°

∠BOC = 180° - 70°

∠BOC = 110°

Hence, the value of ∠BOC is 110°

**Question 4: If the angles of a triangle are in the ratio 2:1:3, then find the measure of smallest angle.**

**Solution:**

Let us assumed that,

Angles of a triangles are 2x, x and 3x, where x is the smallest angle.

We know that, Sum of angles of a triangle is 180°

2x + x + 3x = 180° (angle sum property)

6x = 180°

x = 30°.

Hence, the value of angle smallest angle is 30°.

**Question 5: If the angles A, B and C of △ABC satisfy the relation B – A = C – B, then find the measure of ∠B.**

**Solution:**

We know that, Sum of angles of a triangle is 180°

∠A + ∠B + ∠C = 180°_____(1) (Angle sum property)

It is given that,

∠B – ∠A = ∠C – ∠B

2∠B = ∠C + ∠A ______(2)

Put the value of C + A in equation (1)

2∠B + ∠B = 180°

3∠B =180°

3∠B =180°

∠B = 60°

Hence, the value of ∠B is 60°.