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JEE Mains · Maths · STD 11 - Trigonometrical equations

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes \(18\)  min, for the angle of depression of the car to change from \(30^o\) to \(45^o,\) then after this, the time taken (in min) by the car to reach the foot of the tower, is

  1. A \(9(1 + \sqrt 3 )\)
  2. B \(\frac{9}{2}(\sqrt 3  - 1)\)
  3. C \(18(1 + \sqrt 3 )\)
  4. D \(18( \sqrt 3 -1 )\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(9(1 + \sqrt 3 )\)

Step-by-step Solution

Detailed explanation

Here ; \(\angle DOA\, = \,{45^o}\,;\,\angle DOB\, = \,{60^o}\) Now, let height of tower \(=\,h\) In \(\Delta DOA\,\), \(\tan (\angle DOA)\, = \,\frac{{DA}}{{OD}}\) \( \Rightarrow \,\tan \,{45^o}\, = \frac{{DA}}{h}\, \Rightarrow \,h = \,DA\) Now, in \(\Delta DOB\)…
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