maths----geometry

alet ∠ BDC = β, then ∠ BDA = 180° - βand AD = DC = x consider △ BDC , cos β = (4^2 + x^2 - 4^2 )/2(4)(x) = x^2/8xconsider △ BDA, cos (180° - β) = (4^2 + x^2 - 8^2 )/2(4)(x) = (x^2 - 48)/8x cos β + cos (180° - β) = x^2/8x + (x^2 - 48)/8xcos β - cos β = (2x^2 - 48)/8xtherefore 2x^2 - 48 = 0x^2 = 24 , x =2√6AC = 2x = 4√ 6 cm barea of △ DBC = 1/2 (4)(4)sin Φ = 8sin Φarea of △ ABC = 1/2 (4)(8) sin(Φ+θ) = 16 sin(Φ+θ)since area of △ ADB and area of △ CDB are equal (equal base ,equal height)area of △ DBC x 2 = area of △ ABC 8sin Φ x 2 = 16 sin(Φ+θ)sin(Φ+θ) = sinΦ csin(Φ+θ) = sinΦ (ANSWER b) = sin(180° - Φ)therefore (Φ+θ) = 180° - Φ 2Φ+θ = 180°