A wire is attached from a point $A$ on the ground to the top of a pole $BC$, making an angle of elevation as $60^\circ$. If $AB = 5\sqrt{3}\text{ m}$, then length of the wire is
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In a $30^\circ-60^\circ-90^\circ$ right triangle, the hypotenuse is always twice the length of the side adjacent to the $60^\circ$ angle.
Since $AB = 5\sqrt{3}\text{ m}$, the hypotenuse $AC$ is immediately $2 \times 5\sqrt{3} = 10\sqrt{3}\text{ m}$.
Step 1: Recognise the special triangle.
In right triangle $ABC$ with $\angle B = 90^\circ$ and the angle of elevation $\angle A = 60^\circ$, the remaining angle at $C$ is $30^\circ$, so this is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Step 2: Use the fixed side ratio of this triangle.
In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, the hypotenuse is always exactly twice the side adjacent to the $60^\circ$ angle. Here that adjacent side is $AB = 5\sqrt{3}\text{ m}$. Step 3: Compute the wire length.
So the wire $AC = 2 \times AB = 2 \times 5\sqrt{3} = 10\sqrt{3}\text{ m}$.
\[ \boxed{AC = 10\sqrt{3}\text{ m}} \]
