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JEE Mains · Maths · STD 11 - Trigonometrical equations

The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be \(45^o\) from a point \(A\) on the plane. Let \(B\) be the point \(30\, m\) vertically above the point \(A\). If the angle of elevation of the top of the tower from \(B\) be \(30^o\), then the distance (in \(m\)) of the foot of the tower from the point \(A\) is:

  1. A \(15\left( {1 + \sqrt 3 } \right)\)
  2. B \(15\left( {3 - \sqrt 3 } \right)\)
  3. C \(15\left( {3 + \sqrt 3 } \right)\)
  4. D \(15\left( {5 - \sqrt 3 } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(15\left( {3 + \sqrt 3 } \right)\)

Step-by-step Solution

Detailed explanation

\(AB = 30\,m\, = NP\) In \(\Delta ANM\,\,\tan {45^o}\, = \frac{{MN}}{{AN}}\, = 1\) \( \Rightarrow \,MN = AN\) \(PM = MN - 30\) \( = AN - 30\) In \(\Delta BPM\,\,\tan \,{30^o}\, = \,\frac{{PM}}{{PB}}\, = \,\frac{{AN - 30}}{{AN}}\)…
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