JEE Mains · Maths · STD 11 - Trigonometrical equations
The angle of elevation of the top of a vertical tower from a point \(A\), due east of it is \(45^o\) . The angle of elevation of the top of the same tower from a point \(B\). due south of \(A\) is \(30^o\). If the distance between \(A\) and \(B\) is \(54\sqrt 2 \,m\), then the height of the tower (in metres), is
- A \(108\)
- B \(36\sqrt 3 \)
- C \(54\sqrt 3 \)
- D \(54\)
Answer & Solution
Correct Answer
(D) \(54\)
Step-by-step Solution
Detailed explanation
Let \(AP=x\) \(BP=y\) \(\tan 45^o\,=\,\frac {H}{x}\Rightarrow H=x\) \(\tan 30^o\,=\,\frac {H}{y}\Rightarrow y=\sqrt 3H\) \({x^2}\, + \,{(54\sqrt 2 )^2}\, = {y^2}\) \({H^2}\, + \,{(54\sqrt 2 )^2}\, = 3{H^2}\) \({(54\sqrt 2 )^2}\, = 2{H^2}\) \(54\sqrt 2 \, = \,\sqrt 2 H\)…
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