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4. (there is a right - angled triangle in the picture. one of the acute…

Question

4.
(there is a right - angled triangle in the picture. one of the acute angles is 30°, the hypotenuse length is 14√3, the length of the side opposite the 30° angle is y, and the length of the other right - angled side is x.)

Explanation:

Step1: Identify triangle type

This is a right - triangle with one angle \(30^{\circ}\), so the other non - right angle is \(60^{\circ}\) (since the sum of angles in a triangle is \(180^{\circ}\), \(180 - 90-30 = 60^{\circ}\)). Let's assume the side opposite \(30^{\circ}\) is \(y\), the side opposite \(60^{\circ}\) is \(x\), and the hypotenuse is \(14\sqrt{3}\).

Step2: Use trigonometric ratios

We know that in a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)

For angle \(30^{\circ}\):

  • \(\sin30^{\circ}=\frac{y}{14\sqrt{3}}\), and \(\sin30^{\circ}=\frac{1}{2}\), so \(\frac{1}{2}=\frac{y}{14\sqrt{3}}\), then \(y = \frac{14\sqrt{3}}{2}=7\sqrt{3}\)
  • \(\cos30^{\circ}=\frac{x}{14\sqrt{3}}\), and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\)

Cross - multiply: \(x=\frac{\sqrt{3}\times14\sqrt{3}}{2}=\frac{14\times3}{2}=21\)

Answer:

If we want to find \(x\), \(x = 21\); if we want to find \(y\), \(y=7\sqrt{3}\)