From the top of a 20 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is at 45°, then the height of the tower is \( \sqrt{3} = 1.732 \):
54.64 m
Let the height of the building be AB = 20 m. Let the height of the tower be CD = h m. The angle of elevation of the top of the tower from the top of the building is 60°, so ∠EAB = 60°. The angle of depression of the foot of the tower from the top of the building is 45°, so ∠DAB = 45°. In right-angled triangle ABE, we have:
\(\tan 60° = \frac{BE}{AB}\)
\(\sqrt{3} = \frac{BE}{20}\)
\(BE = 20\sqrt{3}\)
In right-angled triangle ABD, we have:
\(\tan 45° = \frac{AB}{BD}\)
\(1 = \frac{20}{BD}\)
\(BD = 20\)
Since BD = AE = 20 m, the height of the tower is given by:
\(CD = CE + ED = CE + AB\)
\(h = 20\sqrt{3} + 20\)
\(h = 20(\sqrt{3} + 1)\)
\(h = 20(1.732 + 1) = 20(2.732) = 54.64 m\)
Therefore, the height of the tower is 54.64 m.
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