Question

find q.

write your answer in simplest radical form.
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Explanation:

Step1: Use the sine function

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 60^{\circ}\), the opposite side is \(\sqrt{6}\) and the hypotenuse is \(q\). So, \(\sin60^{\circ}=\frac{\sqrt{6}}{q}\).
Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{q}\).

Step2: Solve for \(q\)

Cross - multiply: \(\sqrt{3}q = 2\sqrt{6}\).
Then \(q=\frac{2\sqrt{6}}{\sqrt{3}}\).
Rationalize the denominator: \(q = 2\sqrt{\frac{6}{3}}\).
Simplify \(\frac{6}{3}=2\), so \(q = 2\sqrt{2}\).

Answer:

\(2\sqrt{2}\)