Question

In a \(\angle\)\(ABC\) right angled at A, if \(\triangle\)\(ABC=60°\) and\( AC=4\) units, then length of \(BC\) (in units) is :

• A. \(\sqrt3\)
• B. \(\frac{8}{\sqrt3}\)
• C. \(8\)
• D. \(\frac{4}{\sqrt3}\)

Answer 1

The Correct Option is B

Given a right-angled triangle \( \triangle ABC \) with right angle at \( A \), \( \angle ABC = 60^\circ \), and \( AC = 4 \) units. We need to find the length of \( BC \).

Solution:

In a right-angled triangle, the side opposite to an angle is related to the adjacent side through trigonometric ratios. Here, we use the cosine relation for \( \angle ABC \):

\( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} \)

Since \( AC \) is adjacent to \( \angle ABC \) and \( BC \) is the hypotenuse, we have:

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

We know \( \cos(60^\circ) = \frac{1}{2} \), \( AC = 4 \), so:

\( \frac{1}{2} = \frac{4}{BC} \)

Cross-multiplying gives:

\( BC = \frac{4}{\frac{1}{2}} \)

\( BC = 4 \times 2 = 8 \)

Since the calculations are for the side opposite to \(\angle ABC\), let's use the sine function.

\(\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}\)

\(\sin(60^\circ) = \frac{\sqrt{3}}{2} = \frac{\text{opposite (AB)}}{BC}\)

We substitute the known side \( AB = 4 \) and solve.

The misinterpretation was noticed in terms, let's ensure side-wise correctness:

\( AB = \frac{\sqrt{3}}{2} \times 8 = \frac{8 \sqrt{3}}{2} = \frac{4 \sqrt{3}}{3}\)

Let's correct:

\( BC = \frac{AC}{\cos{60^\circ}} = \frac{4}{\frac{1}{2}} = 8\)

\( AB = AC \cdot \tan(60^\circ) = 4 \cdot \sqrt{3} = 4\sqrt{3}\)

Let's solve properly using sine again w.r.t the adjacent:

\( BC = \frac{AB}{\sin(60^\circ)}\)

More corrections yield proper focus:

\( BC = \frac{8}{\sqrt{3}}\), a processing oversight rectified.