Question

angles of a triangle ABC are in AP and it is being given that b :c = square root 3: square root to find angle A

Answer 1

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