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The angle of elevation of the top of a building from the foot of the tower is $30^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is 60 m high, find the height of the building and distance between the building and the tower. (Use $\sqrt{3} = 1.73$)
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Solution: For correct figure:1 (1 Mark)
Let AB be the building and CD be the tower.
In $\triangle DCA$, $\tan 60^{\circ} = \frac{60}{x}$ (1 Mark)
$\Rightarrow x = 20\sqrt{3} = 34.6$ m (1 Mark)
In $\triangle BAC$, $\tan 30^{\circ} = \frac{h}{x}$ (1 Mark)
$\Rightarrow h = 20$ m (1 Mark)
Let AB be the building and CD be the tower.
In $\triangle DCA$, $\tan 60^{\circ} = \frac{60}{x}$ (1 Mark)
$\Rightarrow x = 20\sqrt{3} = 34.6$ m (1 Mark)
In $\triangle BAC$, $\tan 30^{\circ} = \frac{h}{x}$ (1 Mark)
$\Rightarrow h = 20$ m (1 Mark)