Question : A person 1.8 metres tall is $30 \sqrt{3}$ metres away from a tower. If the angle of elevation from his eye to the top of the tower is 30°, then what is the height (in m) of the tower?
Option 1: 32.5
Option 2: 37.8
Option 3: 30.5
Option 4: 31.8
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Correct Answer: 31.8
Solution : Given: A person 1.8 metres tall is $30 \sqrt{3}$ metres away from a tower. The angle of elevation from his eye to the top of the tower is $30^{\circ}$. We know the formulas, $\tan 30^{\circ}=\frac{1}{\sqrt3}$ and $\tan \theta=\frac{\text{Perpendicular}}{\text{Base}}$. Let the tower's height and the person's height be AB and DE, respectively. In $\triangle ACD$, $\frac{AC}{DC}=\tan 30^{\circ}$ ⇒ $\frac{x}{30\sqrt3}=\frac{1}{\sqrt3}$ ⇒ $x=30$ m The height of the tower AB = 30 + 1.8 = 31.8 m Hence, the correct answer is 31.8 m.
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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:
Question : Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45°, respectively. If the height of the tower is 50 metres, the distance between the two men is: (take $\sqrt{3}=1.732$)
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