Question

find the value of x and y. triangle image with right angle, 60° angle, x as one leg, 9 as the other leg, y as hypotenuse x space x = space y

Explanation:

Step1: Identify triangle type

This is a right - angled triangle with one angle \(60^{\circ}\), so the third angle is \(30^{\circ}\). In a \(30 - 60 - 90\) right - angled triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite to \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\) respectively). The side of length \(9\) is opposite the \(60^{\circ}\) angle, and \(x\) is opposite the \(30^{\circ}\) angle, \(y\) is the hypotenuse.
We know that \(\tan(60^{\circ})=\frac{9}{x}\), and \(\tan(60^{\circ}) = \sqrt{3}\). So \(\sqrt{3}=\frac{9}{x}\), which gives \(x=\frac{9}{\sqrt{3}} = 3\sqrt{3}\) (rationalizing the denominator: \(\frac{9}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)).

Step2: Find the hypotenuse \(y\)

We can use the Pythagorean theorem \(x^{2}+9^{2}=y^{2}\), or use the ratio of sides in a \(30 - 60 - 90\) triangle. In a \(30 - 60 - 90\) triangle, the hypotenuse is twice the length of the side opposite the \(30^{\circ}\) angle. Since \(x\) is opposite \(30^{\circ}\), \(y = 2x\). Substituting \(x = 3\sqrt{3}\), we get \(y=6\sqrt{3}\)? Wait, no, wait. Wait, the side of length \(9\) is opposite \(60^{\circ}\), the side opposite \(30^{\circ}\) is \(x\), and hypotenuse is \(y\). The ratio of sides: side opposite \(30^{\circ}\): side opposite \(60^{\circ}\): hypotenuse \(=1:\sqrt{3}:2\). So if side opposite \(60^{\circ}\) is \(9\) (i.e., \(\sqrt{3}\) part), then the side opposite \(30^{\circ}\) ( \(x\)) is \(\frac{9}{\sqrt{3}}=3\sqrt{3}\) (dividing by \(\sqrt{3}\) to get the \(1\) part), and the hypotenuse \(y\) is \(\frac{9}{\sqrt{3}}\times2=\frac{18}{\sqrt{3}} = 6\sqrt{3}\)? Wait, no, another way: \(\cos(60^{\circ})=\frac{x}{y}\) and \(\sin(60^{\circ})=\frac{9}{y}\). Since \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{9}{y}\), so \(y=\frac{18}{\sqrt{3}}=6\sqrt{3}\) (rationalizing: \(\frac{18}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{18\sqrt{3}}{3} = 6\sqrt{3}\)). And \(\tan(60^{\circ})=\frac{9}{x}\), \(\tan(60^{\circ})=\sqrt{3}\), so \(x = \frac{9}{\sqrt{3}}=3\sqrt{3}\) as before.
Wait, also, using Pythagorean theorem: \(x^{2}+9^{2}=y^{2}\). Substitute \(x = 3\sqrt{3}\), \((3\sqrt{3})^{2}+81=y^{2}\), \(27 + 81=y^{2}\), \(108=y^{2}\), \(y=\sqrt{108}=6\sqrt{3}\) (since \(y>0\)).

Answer:

\(x = 3\sqrt{3}\), \(y = 6\sqrt{3}\)