Question

1. find the missing variables.

hints

1. step 1: divide 24 by $\sqrt{3}$

Explanation:

Step1: Calculate \( y \)

In a \( 30^{\circ}-60^{\circ}-90^{\circ} \) triangle, if the side adjacent to the \( 30^{\circ} \) angle is \( 24 \), and \( y \) is the side opposite the \( 30^{\circ} \) angle.
We know that \(\tan30^{\circ}=\frac{y}{24}\), and \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\). So \( y = 24\times\frac{1}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\).
Another way (using the hint - relation in \( 30 - 60-90 \) triangle: if the side adjacent to \( 30^{\circ} \) is \( a = 24 \), and \( y\) (opposite to \( 30^{\circ} \)) \(y=\frac{24}{\sqrt{3}} = 8\sqrt{3}\).

Step2: Calculate \( x \)

In a \( 30^{\circ}-60^{\circ}-90^{\circ} \) triangle, the hypotenuse \( x \) and the side adjacent to \( 30^{\circ} \) (let's call it \( a = 24 \)) are related by \(\cos30^{\circ}=\frac{24}{x}\), and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\). So \( x=\frac{24}{\cos30^{\circ}}=\frac{24}{\frac{\sqrt{3}}{2}}=\frac{48}{\sqrt{3}} = 16\sqrt{3}\).
Another way: In a \( 30 - 60-90 \) triangle, if the side adjacent to \( 30^{\circ} \) is \( 24 \), and the hypotenuse \( x = 2y\) (since \( y = 8\sqrt{3}\)), then \(x = 16\sqrt{3}\).

Answer:

\(x = 16\sqrt{3}\), \(y=8\sqrt{3}\)