Question

find c.
right triangle with angles 30°, 60°, 90°; vertical leg is √6 mi, horizontal leg is c
write your answer in simplest radical form.
blank miles

Explanation:

Step1: Identify the trigonometric ratio

In a right - triangle, we know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 60^{\circ}\), the opposite side to \(\theta\) is \(\sqrt{6}\) miles and the adjacent side is \(c\). So, \(\tan(60^{\circ})=\frac{\sqrt{6}}{c}\).

Step2: Recall the value of \(\tan(60^{\circ})\)

We know that \(\tan(60^{\circ})=\sqrt{3}\). Substituting this value into the equation from Step 1, we get \(\sqrt{3}=\frac{\sqrt{6}}{c}\).

Step3: Solve for \(c\)

To solve for \(c\), we can cross - multiply. So, \(c\times\sqrt{3}=\sqrt{6}\). Then, \(c = \frac{\sqrt{6}}{\sqrt{3}}\).
We can simplify \(\frac{\sqrt{6}}{\sqrt{3}}\) by using the property \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) (where \(a = 6\) and \(b = 3\)). So, \(\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2}\). Another way is to rationalize or simplify the square roots: \(\frac{\sqrt{6}}{\sqrt{3}}=\frac{\sqrt{3}\times\sqrt{2}}{\sqrt{3}}=\sqrt{2}\) (canceling out \(\sqrt{3}\) from the numerator and the denominator).

Answer:

\(\sqrt{2}\)