Question : From the top of a 20 metres high building, the angle of elevation from the top of a tower is 60° and the angle of depression of its foot is at 45°, then the height of the tower is: $(\sqrt{3} = 1.732)$
Option 1: 45.46 metres
Option 2: 45.64 metres
Option 3: 54.64 metres
Option 4: 54.46 metres
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Correct Answer: 54.64 metres
Solution :
In $\triangle BDE$,
$\tan 45°=\frac{BD}{DE}$
⇒ $1=\frac{20}{DE}$
⇒ $DE=20$ metres
⇒ $DE=BC=20$ metres
In $\triangle ABC$,
$\tan 60°=\frac{AC}{BC}$
⇒ $\sqrt{3}=\frac{AC}{20}$
⇒ $AC=20\sqrt{3}=20\times1.732=34.64$ metres
So, the height of the tower is (34.64 + 20) = 54.64 metres
Hence, the correct answer is 54.64 metres.
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