JEE Mains · Maths · STD 11 - Trigonometrical equations
\(ABC\) 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 \({\cot ^{ - 1}}\left( {3\sqrt 2 } \right)\) and \(\cos e{c^{ - 1}}\left( {2\sqrt 2 } \right)\) respectively, then the height of the tower (in metres) is
- A \(25\)
- B \(10\sqrt 5 \)
- C \(\frac{{100}}{{3\sqrt 3 }}\)
- D \(20\)
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
\(\cos ec\,\beta \, = \,2\sqrt 2 \) \(\cot \,\alpha \, = 3\sqrt 2 \) \(\frac{x}{h}\, = \,3\sqrt 2 \,....\,\,(i)\) So \(\frac{\alpha }{{\sqrt {{{10}^4} - {x^2}} }} = \frac{1}{{\sqrt 7 }}\,....\,(ii)\) From \((i)\) and \((ii)\) \(h\, = \,20\)
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