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