1.

1). 2 – √32).3).4).

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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