Question

find the value of x and y.
(there is a right triangle with a right angle at the bottom left, one angle of 60° at the top left, hypotenuse length 6, vertical leg labeled x, horizontal leg labeled y)

Explanation:

Step1: Identify the triangle type

This is a right - angled triangle with one angle of \(60^{\circ}\), so it's a \(30 - 60 - 90\) triangle. In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest one, opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest, and the hypotenuse is twice the shortest.

First, find the angle opposite to side \(x\). The right angle is \(90^{\circ}\), one angle is \(60^{\circ}\), so the third angle is \(180 - 90 - 60=30^{\circ}\). The hypotenuse is \(6\). The side \(x\) is adjacent to the \(60^{\circ}\) angle and opposite to the \(30^{\circ}\) angle. In a \(30 - 60 - 90\) triangle, the side opposite \(30^{\circ}\) (which is \(x\)) is half of the hypotenuse.

Step2: Calculate \(x\)

Since the hypotenuse \(h = 6\) and the side opposite \(30^{\circ}\) ( \(x\)) is \(\frac{h}{2}\), so \(x=\frac{6}{2}=3\).

Step3: Calculate \(y\)

The side \(y\) is opposite the \(60^{\circ}\) angle. In a \(30 - 60 - 90\) triangle, the side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the side opposite \(30^{\circ}\). We know the side opposite \(30^{\circ}\) is \(x = 3\), so \(y=x\sqrt{3}=3\sqrt{3}\). We can also use trigonometric ratios. Using \(\cos(60^{\circ})=\frac{x}{6}\), since \(\cos(60^{\circ})=\frac{1}{2}\), then \(\frac{1}{2}=\frac{x}{6}\), so \(x = 3\). And using \(\sin(60^{\circ})=\frac{y}{6}\), since \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), then \(y = 6\times\frac{\sqrt{3}}{2}=3\sqrt{3}\).

Answer:

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