Question

find x and y

Explanation:

Step1: Use trigonometric ratios for the first triangle

In the first right - triangle with angle \(A = 30^{\circ}\) and adjacent side \(AC = 3\).
For \(x\) (opposite side to \(30^{\circ}\) angle), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), but for a \(30 - 60-90\) triangle, the side opposite \(30^{\circ}\) is half the hypotenuse. Also, \(\sin30^{\circ}=\frac{x}{y}\) and \(\cos30^{\circ}=\frac{3}{y}\). But using the property of \(30 - 60 - 90\) triangle: if the side adjacent to \(30^{\circ}\) (let \(AC\)) is \(a = 3\), then the hypotenuse \(y\) (by \(\cos30^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}\)), \(y=\frac{3}{\cos30^{\circ}}=\frac{3}{\frac{\sqrt{3}}{2}} = 2\sqrt{3}\). And \(x=\frac{y}{2}=\sqrt{3}\) (from \(30 - 60-90\) triangle ratio \(1:\sqrt{3}:2\))

Step2: Use trigonometric ratios for the second triangle

In the second right - triangle with angle \(A = 30^{\circ}\) and adjacent side \(AC = 12\).
For \(x\) (opposite side to \(30^{\circ}\) angle), using \(\tan30^{\circ}=\frac{x}{12}\), \(x = 12\times\tan30^{\circ}=12\times\frac{1}{\sqrt{3}} = 4\sqrt{3}\). For \(y\) (hypotenuse), using \(\cos30^{\circ}=\frac{12}{y}\), \(y=\frac{12}{\cos30^{\circ}}=\frac{12}{\frac{\sqrt{3}}{2}}=8\sqrt{3}\)

Answer:

For the first triangle: \(x = \sqrt{3}\), \(y = 2\sqrt{3}\); For the second triangle: \(x = 4\sqrt{3}\), \(y = 8\sqrt{3}\)