JEE Mains · Maths · STD 11 - Trigonometrical equations
Let a vertical tower \(AB\) of height \(2 h\) stands on a horizontal ground. Let from a point \(P\) on the ground a man can see upto height \(h\) of the tower with an angle of elevation \(2 \alpha\). When from \(P\), he moves a distance \(d\) in the direction of \(\overline{A P}\), he can see the top B of the tower with an angle of elevation \(\alpha\). If \(d=\sqrt{7} h\), then \(\tan \alpha\) is equal to.
- A \(\sqrt{5}-2\)
- B \(\sqrt{3}-1\)
- C \(\sqrt{7}-2\)
- D \(\sqrt{7}-\sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{7}-2\)
Step-by-step Solution
Detailed explanation
\(\tan 2 \alpha=\frac{h}{x}\) and \(\tan \alpha=\frac{2 h}{x+\sqrt{7} h}\) \(\tan \alpha=\frac{2 h}{h \cot 2 \alpha+\sqrt{7} h}\) \(\tan \alpha=\frac{2}{\frac{\left(1-\tan ^{2} \alpha\right)}{2 \tan \alpha}+\sqrt{7}}\) Put \(\tan \alpha=t\,and\, \operatorname{simplify}\)…
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