Question

The angles of a triangle ABC are in A. P. and it is being given that b:c
d it is being given that b:c = root3:root2
find angle A.​

Answer 1

Answer:

∠ A = 75

Explanation:

Given :

Three angles of triangle are in A.P.

Side ratio are given : √ 3 : √ 2

Let say three angles are    x , y ans z

2 ∠ y =∠ x +∠ z  [ Arithmetic mean ]

∠ x + ∠ y + ∠ z = 180

∠ y + 2 ∠ y = 180

3 ∠ y = 180

∠ y = 60

From sine rule

a / sin ∠ A = b / sin ∠ B = c / sin ∠ C

b / c = sin ∠ B / sin ∠ C

√ 3 / √ 2 = sin 60 / sin ∠ C

sin 60 = √ 3

sin  ∠ C = 1 / √ 2

sin  ∠ C = sin 45

∠ C = 45

Now we have ∠ C = 45 and ∠ B = 60 and ∠ A = ?

2 ∠ y =∠ x +∠ z

2 × 60 = 45 + ∠ x

∠ x = 120 - 45

∠ x = 75

Hence we get angle A = 75 .

Answer 2

Since,A,B and C are in A.P.,

A + C = 2B...(i)  

But,

A + B + C = π ⇒

2B + B = π ..[From(i)]

⇒ B = π /3

 

A + C = π − B = π − π /3

=2π 3 

⇒ C = 2π/3 - A

from sine rule, 

sinB/ b  = sinC/ c

⇒ = sinB/ b = sin(2π/3 - A)/c

sin(2π/3 - A) = c/b* sinB

sin(2π/3 - A) = √2/√3 * sin π/3         {∴ b : c = √3 : √2}

sin(2π/3 - A) = √2/√3 * √3/2  = 1/√2 = π /4

2π/3 - A = π /4

A = 2π/3 - π /4 = 5π / 12

C = 2π/3 - 5π / 12 = (8π - 5π)/12

  = 3π / 12 = π / 12