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: Find the height of the pole first using tangent, instead of using cosine directly.
In right triangle $ABC$, $\tan 60^\circ = \frac{BC}{AB}$, so \[ BC = AB \times \tan 60^\circ = 5\sqrt{3} \times \sqrt{3} = 15\text{ m} \]
Step 2: Apply Pythagoras' theorem to find the wire length AC. \[ AC^2 = AB^2 + BC^2 = (5\sqrt{3})^2 + 15^2 = 75 + 225 = 300 \]
Step 3: Take the square root. \[ AC = \sqrt{300} = 10\sqrt{3}\text{ m} \]
\[ \boxed{10\sqrt{3}\text{ m}} \]