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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Question : The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 metres longer than when it was 60°. The height of the tower is:
Question : On the ground, there is a vertical tower with a flagpole on its top. At a point 9 metres away from the foot of the tower, the angles of elevation of the top and bottom of the flagpole are 60° and 30°, respectively. The height of the flagpole is:
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