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. Triangle $ABC$ is right angled at $B$ with the angle of elevation $60^\circ$ at $A$, so it's a standard $30^\circ$-$60^\circ$-$90^\circ$ triangle. Step 2: Recall the fixed side ratio for this triangle. In such a triangle, the hypotenuse is always exactly twice the side adjacent to the $60^\circ$ angle, i.e. $AC = 2 \times AB$. Step 3: Apply it directly. \[ AC = 2 \times 5\sqrt{3} = 10\sqrt{3}\text{ m} \] Step 4: Conclude. The wire's length is $10\sqrt3\text{ m}$, matching option (2). \[ \boxed{10\sqrt{3}\text{ m}} \]
