MCQOPTIONS
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| 1. |
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are \({\rm{co}}{{\rm{t}}^{ - 1}}\left( {3\sqrt 2 } \right){\rm{\;and\;cose}}{{\rm{c}}^{ - 1}}\left( {2\sqrt 2 } \right)\) respectively, then the height of the tower (in meters) is: |
| A. | \(\frac{{100}}{{3\sqrt 3 }}\) |
| B. | \(10\sqrt 5\) |
| C. | 20 |
| D. | 25 |
| Answer» D. 25 | |