The document RD Sharma Solutions -Ex-9.2, Triangle And Its Angles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q1) The exterior angles, obtained on producing the base of a triangle both ways are 104 ^{0} and 136^{0}. Find all the angles of the triangle.**

**Solution:**

âˆ ACD = âˆ ABC + âˆ BAC [Exterior angle property]

Nowâˆ ABC = 180^{0} âˆ’ 136^{0} = 44^{0} [Linera pair]

âˆ ACB = 180^{0} âˆ’ 104^{0} = 76^{0} [Linera pair]

Now,InÎ”ABC

âˆ A + âˆ ABC + âˆ ACB = 180^{0 }[Sum of all angles of a triangle]

â‡’ âˆ A + 44^{0 }+ 76^{0 }= 180^{0}

â‡’ âˆ A = 180^{0} âˆ’ 44^{0 }âˆ’ 76^{0}

â‡’ âˆ A = 60^{0}

**Q2) In a triangle ABC, 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 = 180 ^{0}.**

**Solution:**

Let âˆ ABD = 2x and âˆ ACE = 2y

âˆ ABC = 180^{0 }âˆ’ 2x [Linera pair]

âˆ ACB = 180^{0 }âˆ’ 2y [Linera pair]

âˆ A + âˆ ABC + âˆ ACB = 180^{0 }[Sum of all angles of a triangle]

â‡’ âˆ A + 180^{0}^{ }âˆ’ 2x + 180^{0 }âˆ’ 2y = 180^{0}

â‡’ âˆ’ âˆ A + 2x + 2y = 180^{0}

â‡’ x + y = 90^{0} + âˆ A

Now in Î”BQC x + y + âˆ BQC = 180^{0} [Sum of all angles of a triangle]

â‡’ 90^{0 }+ âˆ A + âˆ BQC = 180^{0}

â‡’ âˆ BQC = 90^{0 }âˆ’ âˆ Aâ€¦.(i)

and we know thatâˆ BPC = 90^{0} + âˆ Aâ€¦.(ii)

Adding (i) and (ii) we get âˆ BPC + âˆ BQC = 180^{0}

Hence proved.

**Q3) In figure 9.30, the sides BC, CA and AB of a triangle ABC have been produced to D, E and F respectively. If âˆ ACD = 105 ^{0} and âˆ EAF = 45^{0}, find all the angles of the triangle ABC.**

**Solution:**

âˆ BAC = âˆ EAF = 45^{0} [Verticallyopposite angles]

âˆ ABC = 105^{0} âˆ’ 45^{0} = 60^{0} [Exterior angle property]

âˆ ACD = 180^{0} âˆ’ 105^{0} = 75^{0} [Linear pair]

** **

**Q4) Compute the value of x in each of the following figures:**

**(i)**

**Solution:**

âˆ BAC = 180^{0} âˆ’ 120^{0} = 60^{0} [Linear pair]

âˆ ACB = 180^{0 }âˆ’ 112^{0} = 68^{0} [Linear pair]

âˆ´ x = 180^{0 }âˆ’ âˆ BAC âˆ’ âˆ ACB

= 180^{0} âˆ’ 60^{0} âˆ’ 68^{0} = 52^{0} [Sum of all angles of a triangle]

** **

**(ii)**

**Solution:**

âˆ ABC = 180^{0} âˆ’ 120^{0 }= 60^{0} [Linear pair]

âˆ ACB = 180^{0 }âˆ’ 110^{0} = 70^{0} [Linear pair]

âˆ´ eâˆ BAC = x = 180^{0} âˆ’ âˆ ABC âˆ’ âˆ ACB

= 180^{0} âˆ’ 60^{0 }âˆ’ 70^{0} = 50^{0} [Sum of all angles of a triangle]

** **

**(iii)**

**Solution:**

âˆ BAE = âˆ EDC = 52^{0} [Alternate angles]

âˆ´âˆ DEC = x = 180^{0} âˆ’ 40^{0} âˆ’ âˆ EDC = 180^{0} âˆ’ 40^{0} âˆ’ 52^{0 }

= 180^{0} âˆ’ 92^{0} = 88^{0} [Sum of all angles of a triangle]

**(iv) **

**Solution:**

CD is produced to meet AB at E.

âˆ BEC = 180^{0} âˆ’ 45^{0 }âˆ’ 50^{0} = 85^{0 }[Sumofall angles of a triangle]

âˆ AEC = 180^{0 }âˆ’ 85^{0 }= 95^{0} [Linear pair]

âˆ´x = 95^{0} + 35^{0} = 130^{0} [Exterior angle property] .

**Q5) In figure 9.35, AB divides âˆ DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.**

**Solution:**

Letâˆ BAD = Z,âˆ BAC = 3Z

â‡’ âˆ BDA = âˆ BAD = Z (âˆµAB = DB)

Nowâˆ BAD + âˆ BAC + 108^{0} = 180^{0} [Linear pair]

â‡’ Z + 3Z + 108^{0 }= 180^{0}

â‡’ 4Z = 72^{0}

â‡’ Z = 18^{0}

Now, In Î”ADCâˆ ADC + âˆ ACD = 108^{0} [Exterior angle property]

â‡’ x + 18^{0} = 180^{0}

â‡’ x = 90^{0}

**Q6) ABC is a triangle. The bisector of the exterior angle at B and the bisector of âˆ C intersect each other at D. Prove that âˆ D = ****âˆ A.**

**Solution:**

Let âˆ ABE = 2x and âˆ ACB = 2y âˆ ABC = 180^{0} âˆ’ 2x [Linear pair]

âˆ´âˆ A = 180^{0} âˆ’ âˆ ABC âˆ’ âˆ ACB [ angle sum property]

= 180^{0 }âˆ’ 180^{0} + 2x + 2y = 2(x âˆ’ y)â€¦..(i)

Now,âˆ D = 180^{0} âˆ’ âˆ DBC âˆ’ âˆ DCB

â‡’ âˆ D = 180^{0} âˆ’ (x + 180^{0} âˆ’ 2x) âˆ’ y

â‡’ âˆ D = 180^{0} âˆ’ x âˆ’ 180^{0} + 2x âˆ’ y = (x âˆ’ y) = âˆ Aâ€¦..from(i)

Hence,âˆ D = âˆ A.

**Q7) In figure 9.36, ACâŠ¥CE and âˆ A:âˆ B:âˆ C = 3:2:1,find**

**Solution:**

âˆ A:âˆ B:âˆ C = 3:2:1

Let the angles be 3x,2x and x

â‡’ 3x + 2x + x = 180^{0} [ angle sum property]

â‡’ 6x = 180^{0}

â‡’ x = 30^{0 }= âˆ ACB

âˆ´âˆ ECD = 180^{0 }âˆ’ âˆ ACB âˆ’ 90^{0 }[Linear pair]

= 180^{0} âˆ’ 30^{0} âˆ’ 90^{0} = 60^{0}

âˆ´âˆ ECD = 60^{0}

** **

**Q8) In figure 9.37, AMâŠ¥BC and AN is the bisector of âˆ A. If âˆ B = 650 and âˆ C = 33 ^{0},find âˆ MAN..**

**Solution:**

Letâˆ BAN = âˆ NAC = x [âˆµAN bisects âˆ A]

âˆ´âˆ ANM = x + 33^{0} [Exterior angle property]

In Î”AMB âˆ BAM = 90^{0 }âˆ’ 65^{0 }= 25^{0} [Exterior angle property]

âˆ´âˆ MAN = âˆ BAN âˆ’ âˆ BAM = (x âˆ’ 25)^{0}

Now in Î”MAN,(x âˆ’ 25)^{0 }+ (x + 33)^{0 }+ 90^{0} = 180^{0 }[ angle sum property]

â‡’ 2x + 8^{0} = 90^{0}

â‡’ 2x = 82^{0}

â‡’ x = 41^{0}

âˆ´MAN = x âˆ’ 25^{0} = 41^{0} âˆ’ 25^{0} = 16^{0}

**Q9) In a triangle ABC, AD bisects âˆ A and âˆ C> âˆ B.Provethat âˆ ADB > âˆ ADC..**

**Solution:**

âˆµâˆ C>âˆ B [Given]

â‡’ âˆ C + x>âˆ B + x [Adding x on both sides]

â‡’ 180Â° â€“ âˆ ADC>180^{0} â€“ âˆ ADB

â‡’ âˆ’âˆ ADC> â€“ âˆ ADB

â‡’ âˆ ADB > âˆ ADC

Hence proved.

**Q10) In triangle ABC, BDâŠ¥AC and CEâŠ¥AB. If BD and CE intersect at O, prove that âˆ BOC = 180 ^{0} âˆ’ âˆ A..**

**Solution:**

In quadrilateral AEOD

âˆ A + âˆ AEO + âˆ EOD + âˆ ADO = 360^{0}

â‡’ âˆ A + 90^{0} + 90^{0 }+ âˆ EOD = 360^{0}

â‡’ âˆ A + âˆ BOC = 180^{0} [âˆµâˆ EOD = âˆ BOC vertically opposite angles]

â‡’ âˆ BOC = 180^{0} âˆ’ âˆ A

**Q11) In figure 9.38, AE bisects âˆ CAD and âˆ B = âˆ C. Prove that AEâˆ¥BC.**

**Solution:**

Letâˆ B = âˆ C = x

Then,âˆ CAD = âˆ B + âˆ C = 2x (exterior angle)

â‡’ âˆ CAD = x

â‡’ âˆ EAC = x

â‡’ âˆ EAC = âˆ C

These are alternate interior angles for the lines AE and BC

âˆ´AEâˆ¥BC

**Q12) In figure 9.39, ABâˆ¥DE.Findâˆ ACD.**

**Solution:**

Since ABâˆ¥DE

âˆ´âˆ ABC = âˆ CDE = 400 [Alternate angles]

âˆ´âˆ ACB = 180^{0} âˆ’ âˆ ABC âˆ’ âˆ BAC

= 180^{0} âˆ’ 40^{0} âˆ’ 30^{0 }= 110^{0}

âˆ´âˆ ACD = 180^{0} âˆ’ 110^{0} [Linear pair] = 70^{0}

** **

**Q13)** **. Which of the following statements are true (T) and which are false (F) :**

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

**Solution: **(i) T

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

**Solution: **(ii) F

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

**Solution: **(iii) F

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

**Solution: **(iv) F

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

**Solution: **(v) T

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

**Solution:** (vi) F

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

**Solution: **(vii) T

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

**Solution:** (viii) T

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

**Solution: **(ix) F

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

**Solution: **(x) T

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

**Solution: **(xi) T

** **

**Q14) 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^{0}

(ii) Interior

(iii) Greater

(iv) One

(v) One