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1). 2 – √32).3).4). |
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Answer» ½ × AC × BD = ½ × AB × BC [Equating area] AC × BD = AB × BC 4 BD × BD = AB × BC [∵ AC = 4 BD] Dividing by BC2 both the sides 4 × BD/BC × BD/BD = AB/BC In triangle BDC, sin C = BD/BC and In triangle ABC, tan C = AB/BC 4sin2C = tan C 4sin2C = sinC/cos C 4sinC cos C = 1 2sin C cos C = ½ Sin2C = ½ [∵ sin 2A = 2SINA cos A] ∴ 2C = 30° and C = 15° tan C = tan 15° To find tan 15° Tan 2θ = 2tanθ/ (1 – tan2 θ) LET tan 15° be x Tan 30° = 2x/(1 – x2) 1/√3 = 2x/(1 – x2) 1 – x2 = 2√3x x2 + 2√3x – 1 = 0 Using quadratic formula and ignoring the negative term x = (-2√3 + 4)/2 = 2 – √3 |
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