Question

15\sqrt{3} 60^{\circ} x

Explanation:

Step1: Recall trigonometric ratio

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here \(\theta = 60^{\circ}\), the side opposite to \(60^{\circ}\) is \(15\sqrt{3}\), and the hypotenuse is \(x\). So \(\sin60^{\circ}=\frac{15\sqrt{3}}{x}\).

Step2: Substitute the value of \(\sin60^{\circ}\)

Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{x}\).

Step3: Solve for \(x\)

Cross - multiply: \(\sqrt{3}x = 2\times15\sqrt{3}\). Divide both sides by \(\sqrt{3}\), we get \(x = 30\).

Answer:

\(30\)