Question

if in a triangle ABC, the angles are in A.P. and b:c = √3:√2, then angle A = ...........​

Answer 1

Answer:

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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

$\boxed{20 \: thanks = inbox}$

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Answer 2

Answer:

B=60°, C=45°, A=180°-105°=75°