Question

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 \(AB = 5\sqrt{3}\) m, then length of the wire is

• A. \(10\) m
• B. \(10\sqrt{3}\) m
• C. \(15\) m
• D. \(\frac{5}{2}\sqrt{3}\) m

Hint:

In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the hypotenuse is always twice the length of the side adjacent to the \(60^{\circ}\) angle.

Answer 1

The Correct Option is B

To solve the problem, we need to find the length of the wire AC. The wire is represented by the hypotenuse of the right triangle ABC, where:

• Angle of elevation, \(\angle CAB = 60^\circ\)
• Base, \(AB = 5\sqrt{3}\) m
• We need to find the hypotenuse, AC

We can use the trigonometric relation for a right triangle:

\(\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}\)

Substitute the known values:

\(\cos 60^\circ = \frac{AB}{AC}\)

We know that \(\cos 60^\circ = \frac{1}{2}\). Thus:

\(\frac{1}{2} = \frac{5\sqrt{3}}{AC}\)

Solving for AC gives:

\(AC = 2 \times 5\sqrt{3} = 10\sqrt{3}\) m

Therefore, the length of the wire is \(10\sqrt{3}\) m, which matches the correct answer.