Discussion
A man on a cliff observes a fishing trawler at an angle of depression of \( 30^{\circ} \), the trawler approaching the shore directly beneath him. Six minutes later the angle of depression is found to be \( 60^{\circ} \). How much longer will the trawler take to reach the shore?
Answer: C
Let the cliff be \( h \). Depression from the top equals elevation from the trawler, so:
\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{60} = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}} \]
In 6 minutes the trawler covers
\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \]
and the distance still to go is \( \dfrac{h}{\sqrt{3}} \), exactly half of that. At constant speed the remaining time is therefore half of 6 minutes:
\[ t = 6 \times \frac{h/\sqrt{3}}{2h/\sqrt{3}} = 3 \text{ minutes} \]
The cliff height cancels, so it is never needed.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development