Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1) The exterior angles, obtained on producing the base of a triangle both ways are 104⁰  and 136⁰. Find all the angles of the triangle.

Solution:

∠ACD = ∠ABC + ∠BAC [Exterior angle property]
Now∠ABC = 180⁰ − 136⁰ = 44⁰ [Linera pair]
∠ACB = 180⁰ − 104⁰ = 76⁰ [Linera pair]
Now,InΔABC
∠A + ∠ABC + ∠ACB = 180⁰ [Sum of all angles of a triangle]
⇒ ∠A + 44⁰ + 76⁰ = 180⁰
⇒ ∠A = 180⁰ − 44⁰ − 76⁰
⇒ ∠A = 60⁰

Q2) 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 = 180⁰.

Solution:

Let ∠ABD = 2x and ∠ACE = 2y
∠ABC = 180⁰ − 2x [Linera pair]
∠ACB = 180⁰ − 2y [Linera pair]
∠A + ∠ABC + ∠ACB = 180⁰ [Sum of all angles of a triangle]
⇒ ∠A + 180⁰ − 2x + 180⁰ − 2y = 180⁰
⇒ − ∠A + 2x + 2y = 180⁰
⇒ x + y = 90⁰ + ∠A

Now in ΔBQC x + y + ∠BQC = 180⁰ [Sum of all angles of a triangle]
⇒ 90⁰ + ∠A + ∠BQC = 180⁰
⇒ ∠BQC = 90⁰ − ∠A….(i)
and we know that∠BPC = 90⁰ + ∠A….(ii)
Adding (i) and (ii) we get ∠BPC + ∠BQC = 180⁰

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 ∠ACD = 105⁰ and ∠EAF = 45⁰, find all the angles of the triangle ABC.

Solution:

∠BAC = ∠EAF = 45⁰ [Verticallyopposite angles]
∠ABC = 105⁰ − 45⁰ = 60⁰ [Exterior angle property]
∠ACD = 180⁰ − 105⁰ = 75⁰ [Linear pair]

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

(i)

Solution:

∠BAC = 180⁰ − 120⁰ = 60⁰ [Linear pair]
∠ACB = 180⁰ − 112⁰ = 68⁰ [Linear pair]
∴ x = 180⁰ − ∠BAC − ∠ACB
= 180⁰ − 60⁰ − 68⁰ = 52⁰ [Sum of all angles of a triangle]

(ii)

Solution:

∠ABC = 180⁰ − 120⁰ = 60⁰ [Linear pair]
∠ACB = 180⁰ − 110⁰ = 70⁰ [Linear pair]
∴ e∠BAC = x = 180⁰ − ∠ABC − ∠ACB
= 180⁰ − 60⁰ − 70⁰ = 50⁰ [Sum of all angles of a triangle]

(iii)

Solution:

∠BAE = ∠EDC = 52⁰ [Alternate angles]
∴∠DEC = x = 180⁰ − 40⁰ − ∠EDC = 180⁰ − 40⁰ − 52⁰ 
= 180⁰ − 92⁰ = 88⁰ [Sum of all angles of a triangle]

(iv) 

Solution:

CD is produced to meet AB at E.

∠BEC = 180⁰ − 45⁰ − 50⁰ = 85⁰ [Sumofall angles of a triangle]
∠AEC = 180⁰ − 85⁰ = 95⁰ [Linear pair]
∴x = 95⁰ + 35⁰ = 130⁰ [Exterior angle property] .

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

Solution:

Let∠BAD = Z,∠BAC = 3Z
⇒ ∠BDA = ∠BAD = Z           (∵AB = DB)
Now∠BAD + ∠BAC + 108⁰ = 180⁰ [Linear pair]
⇒ Z + 3Z + 108⁰ = 180⁰
⇒ 4Z = 72⁰
⇒ Z = 18⁰
Now, In ΔADC∠ADC + ∠ACD = 108⁰ [Exterior angle property]
⇒ x + 18⁰ = 180⁰
⇒ x = 90⁰

Q6) 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 = ∠A.

Solution:

Let ∠ABE = 2x and ∠ACB = 2y ∠ABC = 180⁰ − 2x [Linear pair]
∴∠A = 180⁰ − ∠ABC − ∠ACB [ angle sum property]
= 180⁰ − 180⁰ + 2x + 2y = 2(x − y)…..(i)
Now,∠D = 180⁰ − ∠DBC − ∠DCB
⇒ ∠D = 180⁰ − (x + 180⁰ − 2x) − y
⇒ ∠D = 180⁰ − x − 180⁰ + 2x − y = (x − y) = ∠A…..from(i)
Hence,∠D = ∠A.

Q7) In figure 9.36, AC⊥CE and ∠A:∠B:∠C = 3:2:1,find

Solution:

∠A:∠B:∠C = 3:2:1
Let the angles be 3x,2x and x
⇒ 3x + 2x + x = 180⁰ [ angle sum property]
⇒ 6x = 180⁰
⇒ x = 30⁰ = ∠ACB
∴∠ECD = 180⁰ − ∠ACB − 90⁰ [Linear pair]
= 180⁰ − 30⁰ − 90⁰ = 60⁰
∴∠ECD = 60⁰

Q8) In figure 9.37, AM⊥BC and AN is the bisector of ∠A. If ∠B = 650 and ∠C = 33⁰,find ∠ MAN..

Solution:

Let∠BAN = ∠NAC = x [∵AN bisects ∠A]
∴∠ANM = x + 33⁰ [Exterior angle property]
In ΔAMB ∠BAM = 90⁰ − 65⁰ = 25⁰ [Exterior angle property]
∴∠MAN = ∠BAN − ∠BAM = (x − 25)⁰
Now in ΔMAN,(x − 25)⁰ + (x + 33)⁰ + 90⁰ = 180⁰ [ angle sum property]
⇒ 2x + 8⁰ = 90⁰
⇒ 2x = 82⁰
⇒ x = 41⁰
∴MAN = x − 25⁰ = 41⁰ − 25⁰ = 16⁰

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

Solution:

∵∠C>∠B                               [Given]

⇒ ∠C + x>∠B + x                 [Adding x on both sides]

⇒  180° – ∠ ADC>180^{0} – ∠ ADB

⇒  −∠ ADC> – ∠ ADB

⇒  ∠ ADB > ∠ ADC

Hence proved.

Q10) In triangle ABC, BD⊥AC and CE⊥AB. If BD and CE intersect at O, prove that ∠BOC = 180⁰ − ∠A..

Solution:

In quadrilateral AEOD

∠A + ∠AEO + ∠EOD + ∠ADO = 360⁰
⇒ ∠A + 90⁰ + 90⁰ + ∠EOD = 360⁰
⇒ ∠A + ∠BOC = 180⁰ [∵∠EOD = ∠BOC  vertically opposite angles]
⇒ ∠BOC = 180⁰ − ∠A

Q11) In figure 9.38, AE bisects ∠CAD and ∠B = ∠C. Prove that AE∥BC.

Solution:

Let∠B = ∠C = x
Then,∠CAD = ∠B + ∠C = 2x (exterior angle)
⇒ ∠CAD = x
⇒ ∠EAC = x
⇒ ∠EAC = ∠C

These are alternate interior angles for the lines AE and BC

∴AE∥BC

Q12) In figure 9.39, AB∥DE.Find∠ACD.

Solution:

Since AB∥DE

∴∠ABC = ∠CDE = 400 [Alternate angles]
∴∠ACB = 180⁰ − ∠ABC − ∠BAC
= 180⁰ − 40⁰ − 30⁰ = 110⁰
∴∠ACD = 180⁰ − 110⁰ [Linear pair] = 70⁰

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

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

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 60°.

Solution: (iii) F

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

Solution: (iv) F

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

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) 180⁰

(ii) Interior

(iii) Greater

(iv) One

(v) One