AP EAMCET · Maths · Heights and Distances
A person observes the top of a tower from a point \(A\) on the ground. The elevation of the tower from this point is \(60^{\circ}\). He moves \(60 \mathrm{~m}\) in the direction perpendicular to the line joining \(A\) and base of the tower. The angle of elevation of the tower from this point is \(45^{\circ}\). Then, the height of the tower (in metres) is
- A \(60 \sqrt{\frac{3}{2}}\)
- B \(60 \sqrt{2}\)
- C \(60 \sqrt{3}\)
- D \(60 \sqrt{\frac{2}{3}}\)
Answer & Solution
Correct Answer
(A) \(60 \sqrt{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
In \(\triangle A B D\) \(\Rightarrow h=x \sqrt{3}\) In \(\triangle B A C\) \(\begin{aligned} B C^2 & =A B^2+A C^2 \\ \Rightarrow B C & =\sqrt{x^2+(60)^2} \\ & =\sqrt{x^2+3600} \end{aligned}\) \(\text {In } \triangle C B D, \tan 45^{\circ}=\frac{h}{\sqrt{3600+x^2}}\)…
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