Discussion
From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings
Answer: B
Let the tower be \( T \) m high and \( d \) m away. Two sightings:
\[ \tan b = \frac{T}{d} = \frac{4}{5}, \qquad \tan a = \frac{T - 230}{d} = \frac{5}{12} \]
Subtracting eliminates \( T \):
\[ \frac{230}{d} = \frac{4}{5} - \frac{5}{12} = \frac{48 - 25}{60} = \frac{23}{60} \]
\[ d = 230 \times \frac{60}{23} = 600 \text{ m} \]
\[ T = \frac{4}{5} \times 600 = 480 \text{ m}, \qquad T - 230 = 250 \text{ m} \]
The line joining the two tops is the hypotenuse of a right triangle with legs \( d \) and \( T - 230 \):
\[ \sqrt{600^{2} + 250^{2}} = \sqrt{360000 + 62500} = \sqrt{422500} = 650 \text{ m} \]
600 m is the horizontal distance only; the question asks for the slant distance between the tops.
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