Question

find x. 20 10√2 10 10√3

Explanation:

Step1: Identify the triangle type

This is a \(30^{\circ}-60^{\circ}-90^{\circ}\) right - triangle. In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\). The side opposite the \(30^{\circ}\) angle is the shortest side, the side opposite the \(60^{\circ}\) angle is \(\sqrt{3}\) times the shortest side, and the hypotenuse is twice the shortest side.

Step2: Use the ratio to find \(x\)

The side of length \(10\) is opposite the \(60^{\circ}\) angle. Let the side opposite the \(30^{\circ}\) angle be \(a\), the side opposite the \(60^{\circ}\) angle be \(a\sqrt{3}\), and the hypotenuse be \(2a\).
If \(a\sqrt{3}=10\), then \(a = \frac{10}{\sqrt{3}}\) (not relevant here). But we know that \(x\) is opposite the \(60^{\circ}\) angle and using the trigonometric ratio \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\) (not the best approach here).
Another way: In a \(30 - 60-90\) triangle, if the side opposite \(30^{\circ}\) is \(y\), side opposite \(60^{\circ}\) is \(x\), and hypotenuse \(h\). We know that \(\sin60^{\circ}=\frac{x}{h}\) and \(\sin30^{\circ}=\frac{10}{h}\). Since \(\sin30^{\circ}=\frac{1}{2}\), then \(h = 20\).
Also, \(\sin60^{\circ}=\frac{\sqrt{3}}{2}=\frac{x}{h}\). Substituting \(h = 20\), we get \(x = 10\sqrt{3}\).

Answer:

\(10\sqrt{3}\)