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

Step 1: Understanding the Concept:
The situation forms a right-angled triangle \(\triangle ABC\), where \(BC\) is the pole, \(AB\) is the distance on the ground, and \(AC\) is the wire (hypotenuse).
Step 2: Key Formula or Approach:
Using the cosine ratio in \(\triangle ABC\):
\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} \]
Step 3: Detailed Explanation:
Given:
Distance \(AB = 5\sqrt{3}\) m
Angle \(\angle A = 60^{\circ}\)
Let the length of the wire be \(AC\).
\[ \cos 60^{\circ} = \frac{AB}{AC} \]
\[ \frac{1}{2} = \frac{5\sqrt{3}}{AC} \]
\[ AC = 2 \times 5\sqrt{3} = 10\sqrt{3} \text{ m} \]
Step 4: Final Answer:
The length of the wire is \(10\sqrt{3}\) m.