Question

find f.
triangle with right angle, 60° angle, 30° angle, side opposite 60° is 5√6 cm, side f opposite 30°
write your answer in simplest radical form.
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Explanation:

Step1: Identify triangle type

This is a right - triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest side (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Also, we can use trigonometric ratios. Let's use the tangent function. \(\tan(30^{\circ})=\frac{f}{5\sqrt{6}}\) (since \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), here \(\theta = 30^{\circ}\), opposite side is \(f\), adjacent side is \(5\sqrt{6}\)).
We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\). So we have the equation \(\frac{1}{\sqrt{3}}=\frac{f}{5\sqrt{6}}\).

Step2: Solve for \(f\)

Cross - multiply to get \(f\times\sqrt{3}=5\sqrt{6}\). Then \(f = \frac{5\sqrt{6}}{\sqrt{3}}\).
Simplify \(\frac{\sqrt{6}}{\sqrt{3}}\) using the property \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) (\(a\geq0,b > 0\)). So \(\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2}\).
Then \(f = 5\sqrt{2}\).
We can also use the sine function. In a right - triangle, \(\sin(60^{\circ})=\frac{5\sqrt{6}}{\text{hypotenuse}}\), but we can also use the fact that in a \(30 - 60 - 90\) triangle, if the side adjacent to \(30^{\circ}\) is \(5\sqrt{6}\), the side opposite to \(30^{\circ}\) (which is \(f\)) can be found by \(f = 5\sqrt{6}\times\tan(30^{\circ})\) as we did before.

Answer:

\(5\sqrt{2}\)