Question

determining a tangent ratio
what is the value of \\(\tan(60^{\circ})\\)?
\\(\frac{1}{2}\\)
\\(\sqrt{3}\\)
\\(\frac{\sqrt{3}}{2}\\)
\\(\frac{1}{\sqrt{3}}\\)

Explanation:

Step1: Recall Tangent Definition

In a right triangle, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ to $\theta$. For $60^\circ$ in a 30-60-90 triangle, we can also use known ratios. Alternatively, use the triangle given: $\angle B = 60^\circ$, right-angled at $C$. So, opposite to $60^\circ$ is $AC$, adjacent is $BC = 6$. First, find $AC$ using Pythagoras or 30-60-90 ratios. In 30-60-90, sides are $x$, $x\sqrt{3}$, $2x$. Here, hypotenuse $AB = 12$, so shorter leg $BC = 6$ (since $2x = 12 \implies x = 6$), longer leg $AC = 6\sqrt{3}$.

Step2: Calculate $\tan(60^\circ)$

$\tan(60^\circ) = \frac{\text{opposite to } 60^\circ}{\text{adjacent to } 60^\circ} = \frac{AC}{BC} = \frac{6\sqrt{3}}{6} = \sqrt{3}$.

Answer:

$\boldsymbol{\sqrt{3}}$ (corresponding to the option with $\sqrt{3}$)