Question

find the missing side lengths. *
1 poir
2√3
60°
у
х
x = 3, y = √3
x = 3, y = √6
x = √6, y = 3
x = 3√2/3, y = √3

Explanation:

Step1: Use trigonometric ratios

In a right - angled triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here \(\theta = 60^{\circ}\), hypotenuse \(c = 2\sqrt{3}\), opposite side to \(60^{\circ}\) is \(x\), and adjacent side to \(60^{\circ}\) is \(y\).
\(\sin60^{\circ}=\frac{x}{2\sqrt{3}}\), since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{x}{2\sqrt{3}}\).
Cross - multiply: \(x=\frac{\sqrt{3}\times2\sqrt{3}}{2}\).

$$x=\frac{2\times3}{2}=3$$

Step2: Use \(\cos\) ratio

\(\cos60^{\circ}=\frac{y}{2\sqrt{3}}\), since \(\cos60^{\circ}=\frac{1}{2}\), we have \(\frac{1}{2}=\frac{y}{2\sqrt{3}}\).
Cross - multiply: \(y = \sqrt{3}\)

Answer:

\(x = 3,y=\sqrt{3}\)