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An observer 2 m tall is 10 √ 3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

Answer: B

The angle is measured from the observer's eye, 2 m above the ground, so the trigonometry gives only the part of the tower above eye level.

\[ \tan 30^{\circ} = \frac{x}{10\sqrt{3}} \]

\[ x = 10\sqrt{3} \times \frac{1}{\sqrt{3}} = 10 \text{ m} \]

Adding the observer's own height:

\[ \text{tower} = 10 + 2 = 12 \text{ m} \]

Forgetting to add the 2 m gives 10 m, which is offered as the first distractor. The \( 10\sqrt{3} \) is chosen so the radical cancels exactly.

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