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A circus artist is climbing a (20m) long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is (30^{o}) (see figure).

Question

A circus artist is climbing a \(20m\) long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is \(30^{o}\) (see figure).

Answer

\(10\ m\)
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Solution

In the given figure, AB be the height of the pole and\(AC=20m\) be the length of rope which is tied from the top of the pole.
We have to determine the height \(AB=?\)
\(\ln \Delta ABC,\)
\(\sin 30^{\circ }=\frac {AB}{AC}=\frac {AB}{20}\)
\(\Rightarrow\) \(\frac {1}{2}=\frac {AB}{20}\) [\(\because \sin30^{\circ}=\frac{1}{2}\)]
\(\Rightarrow\) \(AB=\frac {20}{2}=10m\)
Hence, the height of the pole is \(10m\)
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