In a Δ A B C , if ∠ A = ∠ B = 1 2 sin - 1 6 + 1 2 3 + sin - 1 1 3 and length of the side opposite to ∠ C is c = 6 ⋅ 3 1 4 , then the area of Δ A B C is

Step-by-step solution

2 ∠ B = sin - 1 6 + 1 2 3 + sin - 1 1 3 = tan - 1 6 + 1 3 - 2 + tan - 1 1 2 = tan - 1 6 + 1 3 - 2 + 1 2 1 - 6 + 1 3 - 2 · 1 2 = tan - 1 6 + 1 2 + 3 - 2 2 3 - 2 - 6 + 1 = tan - 1 1 2 + 2 + 3 - 2 6 - 2 - 6 - 1 = tan - 1 2 3 + 3 - 3 = tan - 1 3 3 - 3 = tan - 1 - 3 = 2 π 3 According to question ∠ A = ∠ B = 1 2 sin - 1 6 + 1 2 3 + sin - 1 1 3 ⇒ ∠ A = ∠ B = 1 2 2 π 3 ⇒ ∠ A = ∠ B = π 3 ∴ ∠ C = π - ∠ A - ∠ B = π - 2 π 3 = π 3 Since Δ ABC is an equilateral Δ ∴ ar Δ ABC = 3 4 a 2 = 3 4 c 2 = 3 4 · 6 ⋅ 3 1 / 4 2 = 3 4 3 ⋅ 6 ×