# Class 9 RD Sharma Solutions – Chapter 9 Triangles and its Angles- Exercise 9.2

**Question 1: The exterior angles, obtained on producing the base of a triangle both ways are 104Â° and 136Â°. Find all the angles of the triangle.**

**Solution:**

Theorems Used:The exterior angle theorem states that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles. (Exterior Angle Theorem)âˆ ACD = âˆ ABC + âˆ BAC [Exterior Angle Theorem]

Find âˆ ABC:

âˆ ABC + âˆ ABE = 180Â° [Linear pair]

âˆ ABC + 136Â° = 180Â°

âˆ ABC = 44Â°

Find âˆ ACB:

âˆ ACB + âˆ ACD = 180Â° [Linear pair]

âˆ ACB + 1040 = 180Â°

âˆ ACB = 76Â°

Now,

Sum of all angles of a triangle = 180Â°

âˆ A + 44Â° + 76Â° = 180Â°

âˆ A = 180Â° âˆ’ 44Â°âˆ’76Â°

âˆ A = 60Â°

Angles of the triangle are âˆ A = 60Â°, âˆ B = 44Â° and âˆ C = 76Â°

(ans)

**Question 2: In **an** â–³ABC, the internal bisectors of âˆ B and âˆ C meet at P and the external bisectors of âˆ B and âˆ C meet at Q. Prove that âˆ BPC + âˆ BQC = 180Â°.**

**Solution:**

In â–³ABC,

BP and CP are an internal bisector of âˆ B and âˆ C respectively

=> External âˆ B = 180Â° – âˆ B

BQ and CQ are an external bisector of âˆ B and âˆ C respectively.

=> External âˆ C = 180Â° – âˆ C

In triangle BPC,

âˆ BPC + 1/2âˆ B + 1/2âˆ C = 180Â°

âˆ BPC = 180Â° – (âˆ B + âˆ C) …. (1)

In triangle BQC,

âˆ BQC + 1/2(180Â° – âˆ B) + 1/2(180Â° – âˆ C) = 180Â°

âˆ BQC + 180Â° – (âˆ B + âˆ C) = 180Â°

âˆ BPC + âˆ BQC = 180Â° [Using (1)]

(Proved)

**Question 3: In figure, the sides BC, CA and AB of a â–³ABC have been produced to D, E and F respectively. If âˆ ACD = 105Â° and âˆ EAF = 45Â°, find all the angles of the â–³ABC.**

**Solution:**

Theorems Used:(i) The exterior angle theorem states that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles. (Exterior Angle Theorem)

(ii) Sum of a linear angle pair is 180Â°

(iii)Vertically opposite angles are equal.

âˆ BAC = âˆ EAF = 45Â° [Vertically opposite angles]

âˆ ACD = 180Â° â€“ 105Â° = 75Â° [Linear pair]

âˆ ABC = 105Â° â€“ 45Â° = 60Â° [Exterior angle property]

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

**(i)**

**Solution:**

âˆ BAC = 180Â° â€“ 120Â° = 60Â° [Linear pair]

âˆ ACB = 180Â° â€“ 112Â° = 68Â° [Linear pair]

Sum of all angles of a triangle = 1800

x = 180Â° âˆ’ âˆ BAC âˆ’ âˆ ACB

= 180Â° âˆ’ 60Â° âˆ’ 68Â° = 52Â°

(ans)

**(ii)**

**Solution:**

âˆ ABC = 180Â° â€“ 120Â° = 60Â° [Linear pair]

âˆ ACB = 180Â° â€“ 110Â° = 70Â° [Linear pair]

Sum of all angles of a triangle = 180Â°

x = âˆ BAC = 180Â° âˆ’ âˆ ABC âˆ’ âˆ ACB

= 180Â° â€“ 60Â° â€“ 70Â° = 50Â°

(ans)

**(iii)**

**Solution:**

âˆ BAE = âˆ EDC = 52Â° [Alternate angles]

Sum of all angles of a triangle = 180Â°

x = 180Â° â€“ 40Â° â€“ 52Â° = 180Â° âˆ’ 92Â° = 88Â°

(ans)

**(iv)**

**Solution:**

CD is produced to meet AB at E.

âˆ BEC = 180Â° â€“ 45Â° â€“ 50Â° = 85Â° [Sum of all angles of a triangle = 180Â°]

âˆ AEC = 180Â° â€“ 85Â° = 95Â° [Linear Pair]

Now, x = 95Â° + 35Â° = 130Â° [Exterior angle Property]

Answer: x = 130Â°

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

**Solution:**

Let âˆ BAD = y, âˆ BAC = 3y

âˆ BDA = âˆ BAD = y (As AB = DB)

Now,

âˆ BAD + âˆ BAC + 108Â° = 180Â° [Linear Pair]

y + 3y + 108Â° = 180Â°

4y = 72Â°

or y = 18Â°

Now, In Î”ADC

âˆ ADC + âˆ ACD = 108Â° [Exterior Angle Property]

x + 18Â° = 180Â°

x = 90Â°

(ans)

**Question 6: ABC is a triangle. The bisector of the exterior angle B and the bisector of âˆ C intersect each other at D. Prove that âˆ D = (1/2)âˆ A.**

**Solution:**

Let âˆ ABE = 2x and âˆ ACB = 2y

âˆ ABC = 180Â° – 2K [Linear pair]

âˆ A = 180Â° â€” âˆ ABC â€” âˆ ACB [Angle sum property]

= 180Â° -180Â° + 2x – 2y

= 2(x – y)

Now, âˆ D = 180Â° – âˆ DBC – âˆ DCB

âˆ D = 180Â° -(x + 180Â° – 2x) – y

= x – y

= (1/2)âˆ A (Hence Proved)

**Question 7: In the figure**,** AC is perpendicular to CE and âˆ A:âˆ B:âˆ C = 3:2:1 Find âˆ ECD.**

**Solution:**

Given that âˆ A:âˆ B:âˆ C = 3:2:1

Let the angles be 3x, 2x and x.

3x + 2x + x = 180Â° [Angle Sum property]

6x = 180Â°

x = 30Â° = âˆ ACB

Therefore,

âˆ ECD = 180Â° – âˆ ACB – 90Â° [Linear Pair]

= 180Â° – 30Â° – 90Â°

= 60Â°

(ans)

**Question 8: In the figure, AM is perpendicular to BC and AN is the** **bisector of **âˆ **A. If **âˆ **B = 65Â° and **âˆ **C = 33Â° find **âˆ **MAN.**

**Solution:**

Let âˆ BAN = âˆ NAC = x [AN bisects âˆ A]

Therefore, âˆ ANM = x + 33Â° [Exterior angle property]

In â–³AMB,

âˆ BAM = 90Â° – 65Â° = 25Â° [Exterior angle property]

Therefore, âˆ MAN = âˆ BAN – âˆ BAM = x – 25Â°

Now in â–³MAN,

(x – 25Â°) + (x + 33Â°) + 90Â° = 180Â° [Angle sum property]

or, 2x + 8Â° = 90Â°

or x = 41Â°

Therefore, âˆ MAN = 41Â° – 25Â° = 16Â°

(ans)

**Question 9: In a â–³ABC, AD bisects âˆ A and âˆ C > âˆ B. Prove that âˆ ADB > âˆ ADC.**

**Solution:**

Let us assume that âˆ BAD = âˆ CAD = x. [given AD bisects âˆ A]

Given that,

âˆ C > âˆ B

or, âˆ C + x > âˆ B + x [Adding x on both sides]

or, 180Â° – âˆ ADC > 180Â° – âˆ ADB [Angle sum property]

or, – âˆ ADC >- âˆ ADB

or, âˆ ADB > âˆ ADC.

(proved)

**Question 10: In a â–³ABC, BD is perpendicular to AC and CE is perpendicular to AB. If BD and CE intersect at O prove that âˆ BOC = 180Â° -âˆ A.**

**Solution:**

In quadrilateral AEOD,

âˆ A + âˆ AEO + âˆ EOD + âˆ ADO = 360Â°

or, âˆ A + 90Â° + 90Â° + âˆ EOD = 360Â°

or, âˆ A + âˆ BOC = 360Â° – 90Â° – 90Â° [âˆ EOD = âˆ BOC as they are vertically opposite angles]

or, âˆ BOC = 180Â° – âˆ A

(proved)

**Question 11: In the figure, AE bisects âˆ CAD and âˆ B = âˆ C. Prove that AE || BC.**

**Solution:**

Let âˆ B = âˆ C = x

Then,

âˆ CAD = âˆ B + âˆ C = 2x (Exterior Angle)

âˆ CAD/2 = x

âˆ EAC = âˆ C [AE bisects âˆ CAD and âˆ C = x assumed]

These are interior angles for line AE and BC,

Therefore,

AE || BC

(proved)

**Question 12: In the figure, AB || DE. Find âˆ ACD.**

**Solution:**

Since, AB || DE

Therefore,

âˆ ABC = âˆ CDE = 40Â° [Alternate Angles]

âˆ ACB = 180Â° – âˆ ABC – âˆ BAC

= 180Â° – 40Â° – 30Â°

= 110Â°

Therefore,

âˆ ACD = 180Â° – âˆ ACB [Linear Pair]

=70Â°

**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 180Â°. **** **

Answer: [True]

**(ii) A triangle can have two right angles. **

Answer: [False]

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

Answer: [False]

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

Answer: [False]

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

Answer: [True]

**(vi) A triangle can have two obtuse angles***. *

Answer: [False]

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

Answer: [True]

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

Answer: [True]

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

Answer: [False]

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

Answer: [True]

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

Answer: [True]

**Question 14: Fill in the blanks to make the following statements true**

**(i) Sum of the angles of a triangle is _________. **

Answer: 180Â°

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

Answer: Interior

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

Answer: Greater

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

Answer: One

**(v) A triangle cannot have more than ________________________ obtuse angles. **

Answer: One