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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 $30$ 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:
Let $AB$ be the building and $CD$ be the tower.
In $\Delta ACD, \tan 60^\circ = \frac{30}{x} \Rightarrow x = 10\sqrt{3}$ ....(i)
In $\Delta CAB, \tan 30^\circ = \frac{h}{x} \Rightarrow x = h\sqrt{3}$ ....(ii)
Using (i) and (ii) $h = 10, x = 10 \times 1.73 = 17.3$
$\therefore$ Height of the building $= 10$ m and distance between the building and the tower $= 17.3$ m
Let $AB$ be the building and $CD$ be the tower.
In $\Delta ACD, \tan 60^\circ = \frac{30}{x} \Rightarrow x = 10\sqrt{3}$ ....(i)
In $\Delta CAB, \tan 30^\circ = \frac{h}{x} \Rightarrow x = h\sqrt{3}$ ....(ii)
Using (i) and (ii) $h = 10, x = 10 \times 1.73 = 17.3$
$\therefore$ Height of the building $= 10$ m and distance between the building and the tower $= 17.3$ m