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

For an acute angle in a right triangle, $\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$

Step2: Identify sides for $\angle B=60^\circ$

In $\triangle ABC$, right-angled at $C$:

• Opposite side to $60^\circ$: $AC$
• Adjacent side to $60^\circ$: $BC=6$

First calculate $AC$ using Pythagoras:
$$AC = \sqrt{AB^2 - BC^2} = \sqrt{12^2 - 6^2} = \sqrt{144-36} = \sqrt{108} = 6\sqrt{3}$$

Step3: Compute $\tan(60^\circ)$

$$\tan(60^\circ) = \frac{AC}{BC} = \frac{6\sqrt{3}}{6} = \sqrt{3}$$

Answer:

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