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

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

Hint:

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 $2 \times 5\sqrt{3} = 10\sqrt{3}\text{ m}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a vertical pole $BC$ standing on the ground.
A wire is attached from point $A$ on the ground to the top of the pole $C$.
The distance from point $A$ on the ground to the foot of the pole $B$ is $AB = 5\sqrt{3}\text{ m}$.
The angle of elevation of the wire with the ground is $\angle CAB = 60^\circ$.
We need to find the length of the wire, which represents the hypotenuse $AC$ of the right-angled triangle $\Delta ABC$.

Step 2: Key Formula or Approach:
In right-angled triangle $\Delta ABC$ (where $\angle B = 90^\circ$):
- Adjacent side to angle $60^\circ$ is $AB = 5\sqrt{3}\text{ m}$.
- Hypotenuse is the wire $AC$.
The trigonometric ratio relating the adjacent side and hypotenuse is the cosine function:
\[ \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \]

Step 3: Detailed Explanation:

• Apply the cosine ratio to the right-angled triangle $\Delta ABC$:
\[ \cos 60^\circ = \frac{AB}{AC} \]

• Substitute the known values ($\cos 60^\circ = \frac{1}{2}$ and $AB = 5\sqrt{3}$):
\[ \frac{1}{2} = \frac{5\sqrt{3}}{AC} \]

• Cross-multiply to solve for the length of the wire $AC$:
\[ AC = 2 \times 5\sqrt{3} \]
\[ AC = 10\sqrt{3}\text{ m} \]


Step 4: Final Answer:
The length of the wire is $10\sqrt{3}\text{ m}$.