Question : The angle of elevation of the top of a tower from a point on the ground is 30° and moving 70 m towards the tower it becomes 60°. The height of the tower is:
Option 1: $10$ m
Option 2: $\frac{10}{\sqrt{3}}$ m
Option 3: $10\sqrt{3}$ m
Option 4: $35\sqrt{3}$ m
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Correct Answer: $35\sqrt{3}$ m
Solution : Let $AB$ be the tower and $C$ and $D$ be the point of observation. $AB = h$ and CD = 70 m In $\triangle ABD$, $\frac{AB}{BD}=\tan$ 60° ⇒ $\frac{AB}{BD}=\sqrt3$ ⇒ $BD = \frac{AB}{\sqrt3}$ ⇒ $BD=\frac{h}{\sqrt3}$ In $\triangle ABC$, $\frac{AB}{BC}= \tan$ 30° ⇒ $\frac{AB}{BC}= \frac{1}{\sqrt3}$ ⇒ $BC = AB\sqrt3$ ⇒ $BC= h\sqrt3$ Now, $CD = (BC-BD)$ ⇒ $CD = \sqrt3h-\frac{h}{\sqrt3}$ ⇒ $\sqrt3h-\frac{h}{\sqrt3} = 70$ ⇒ $h = 35\sqrt3$ m Hence, the correct answer is $35\sqrt3$ m.
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