Let triangle ABC be isosceles triangle such that AB and AC are of equal lenght. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that ∠AOB = 105°, then AD/BE equals?
Explanation:
In △OBD, ∠OBD = 90 - 75 = 15° In △EBC, ∠ECB = 90 - 15 = 75° Similarly, ∠B = 75° and ∠A = 180 - 75 - 75 = 30°
Area of △ABC = 12×AD×BC = 12×BE×AC
⇒ ADBE = ACBC ...(2)
Area of △ABC = 12×AB×AC×Sin 30° = 12×AB×BC×Sin 75°
⇒ ACBC = Sin 75°Sin 30° = 2 × Sin 75° = 2 × Cos (90 - 15)° = 2 × Cos 15°
⇒ ADBE = ACBC = 2 × Cos 15°
Hence, option (a).
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