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
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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
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
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