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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