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in Trigonometry by (33.4k points)
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ABCD is a square plot. The angle of elevation of the top of the pole standing at D from A or C is 30° and that from B is θ. Then tan θ is equal to

(a) \(\sqrt6\)

(b) \(1/\sqrt6\)

(c) \(\sqrt3/2\)

(d) \(\sqrt2/3\)

1 Answer

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Best answer

(b) \(1/\sqrt6\) 

Let PD be the pole standing at the corner D of the square plot ABCD. 

Let PD = h m (say)

Given,

∠PCD = 30°, ∠DAP = 30° and ∠PBD = θ.

From rt. Δ PAD or Δ PCD PD = AD tan 30° or PD = CD tan 30° 

Let each side of the square ABCD be a metres.

(Then, PD = a tan 30° = \(\frac{a}{\sqrt3}\))

BD being the diagonal of square ABCD,

BD = \(a\sqrt2\)

∴ In rt. Δ PBD,

tan θ = \(\frac{PD}{BD}\) = \(\frac{\frac{a}{\sqrt3}}{\frac{a}{\sqrt2}}\) \(\frac{1}{\sqrt6}\).

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