Question

find the area
x
9
60°
y

Explanation:

Step1: Identify triangle type

This is a right - angled triangle with one angle \(60^{\circ}\), so the other non - right angle is \(30^{\circ}\). In a \(30 - 60 - 90\) right - angled triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest side. Here, the side of length 9 is opposite the \(30^{\circ}\) angle? Wait, no. Wait, the right angle, one angle is \(60^{\circ}\), so the angles are \(90^{\circ},60^{\circ},30^{\circ}\). Let's denote the sides: let the side adjacent to \(60^{\circ}\) (and opposite \(30^{\circ}\)) be \(x\), the side adjacent to \(30^{\circ}\) (and opposite \(60^{\circ}\)) be 9, and the hypotenuse be \(y\). Wait, in a right - angled triangle, \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{9}{x}\), so \(\sqrt{3}=\frac{9}{x}\), so \(x = \frac{9}{\sqrt{3}}=3\sqrt{3}\)? Wait, no, maybe I mixed up. Wait, the right angle, so the two legs are \(x\) and 9, and the angle between \(x\) and the hypotenuse is \(60^{\circ}\). Wait, let's use trigonometry. In the right - angled triangle, \(\cos(60^{\circ})=\frac{x}{y}\), \(\sin(60^{\circ})=\frac{9}{y}\), and \(\tan(60^{\circ})=\frac{9}{x}\). Since \(\tan(60^{\circ})=\sqrt{3}=\frac{9}{x}\), then \(x=\frac{9}{\sqrt{3}} = 3\sqrt{3}\)? Wait, no, \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\), if the angle is \(60^{\circ}\), the opposite side is 9, and the adjacent side is \(x\), then \(\tan(60^{\circ})=\frac{9}{x}\), so \(x = \frac{9}{\tan(60^{\circ})}=\frac{9}{\sqrt{3}}=3\sqrt{3}\). Wait, no, \(\tan(60^{\circ})=\sqrt{3}\), so \(x=\frac{9}{\sqrt{3}} = 3\sqrt{3}\). Then the area of a right - angled triangle is \(A=\frac{1}{2}\times\text{base}\times\text{height}\). The two legs are \(x\) and 9. Wait, no, maybe the legs are \(x\) and 9. Wait, let's re - examine. The triangle is right - angled, so area \(A=\frac{1}{2}\times\text{leg}_1\times\text{leg}_2\). Let's find the lengths of the two legs. Let the angle at the left vertex be \(60^{\circ}\), the right angle is at the top, so the legs are the vertical leg (length 9) and the horizontal leg (length \(x\)). Then, using \(\tan(60^{\circ})=\frac{9}{x}\), so \(x = \frac{9}{\tan(60^{\circ})}=\frac{9}{\sqrt{3}} = 3\sqrt{3}\)? Wait, no, \(\tan(60^{\circ})=\sqrt{3}=\frac{\text{opposite}}{\text{adjacent}}\), if the angle is \(60^{\circ}\), the opposite side is 9, adjacent is \(x\), so \(x=\frac{9}{\sqrt{3}} = 3\sqrt{3}\). Then the area is \(\frac{1}{2}\times x\times9=\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\)? Wait, no, maybe I got the sides wrong. Wait, another approach: in a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\). If the side opposite \(30^{\circ}\) is \(x\), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Here, if the side of length 9 is opposite \(60^{\circ}\), then \(x\sqrt{3}=9\), so \(x = \frac{9}{\sqrt{3}}=3\sqrt{3}\) (opposite \(30^{\circ}\)), and the other leg (opposite \(90^{\circ}\) is not, wait, no, the legs are opposite \(30^{\circ}\) and \(60^{\circ}\), and the hypotenuse is opposite \(90^{\circ}\). Wait, I think I made a mistake. Let's start over. The triangle is right - angled, so area \(A=\frac{1}{2}\times a\times b\), where \(a\) and \(b\) are the two legs. Let the angle of \(60^{\circ}\) be one of the acute angles. So, \(\sin(60^{\circ})=\frac{\text{opposite leg}}{\text{hypotenuse}}\), \(\cos(60^{\circ})=\frac{\text{adjacent leg}}{\text{hypotenuse}}\), \(\tan(60^{\circ})=\frac{\text{opposite leg}}{\text{adjacent leg}}\). Let's assume that the leg with length 9 is opposite the \(60^{\circ}\) angle, and the other leg (let's call it \(x\)) is adjacent to the \(60^{\circ}\) angle. Then \(\tan(60^{\circ})=\frac{9}{x}\), so \(x=\frac{9}{\tan(60^{\circ})}=\frac{9}{\sqrt{3}} = 3\sqrt{3}\). Then the area is \(\frac{1}{2}\times x\times9=\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\approx\frac{27\times1.732}{2}\approx23.38\). Wait, but maybe the leg of length 9 is adjacent to the \(60^{\circ}\) angle. Then \(\tan(60^{\circ})=\frac{\text{opposite leg}}{9}\), so opposite leg \(=9\times\tan(60^{\circ}) = 9\sqrt{3}\). Then area is \(\frac{1}{2}\times9\times9\sqrt{3}=\frac{81\sqrt{3}}{2}\), which is too big. Wait, the angle is \(60^{\circ}\), and it's a right - angled triangle, so the angles are \(90^{\circ},60^{\circ},30^{\circ}\). In a \(30 - 60 - 90\) triangle, the side opposite \(30^{\circ}\) is the shortest side. So if the side of length 9 is opposite \(30^{\circ}\), then the side opposite \(60^{\circ}\) is \(9\sqrt{3}\), and the hypotenuse is \(18\). But then the area would be \(\frac{1}{2}\times9\times9\sqrt{3}=\frac{81\sqrt{3}}{2}\). But that seems inconsistent. Wait, maybe the triangle is a right - angled triangle with legs \(x\) and 9, and angle between \(x\) and the hypotenuse is \(60^{\circ}\). Then, using trigonometry, \(\cos(60^{\circ})=\frac{x}{y}\) (hypotenuse \(y\)), \(\sin(60^{\circ})=\frac{9}{y}\). Since \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}=\frac{9}{y}\), then \(y=\frac{18}{\sqrt{3}} = 6\sqrt{3}\). Then \(x = y\cos(60^{\circ})=6\sqrt{3}\times\frac{1}{2}=3\sqrt{3}\). Then area \(A=\frac{1}{2}\times x\times9=\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\approx23.38\). Wait, now I'm confused. Wait, let's use the formula for the area of a right - angled triangle: \(A=\frac{1}{2}ab\), where \(a\) and \(b\) are the legs. Let's find the legs. The triangle has angles \(90^{\circ},60^{\circ},30^{\circ}\). Let's say the side of length 9 is opposite the \(30^{\circ}\) angle. Then the side opposite \(60^{\circ}\) is \(9\sqrt{3}\), and the hypotenuse is \(18\). Then the legs are 9 and \(9\sqrt{3}\), so area \(A=\frac{1}{2}\times9\times9\sqrt{3}=\frac{81\sqrt{3}}{2}\approx70.15\). Wait, now I see my mistake. The angle of \(60^{\circ}\): if the right angle is at the top, and the angle at the bottom left is \(60^{\circ}\), then the side opposite \(60^{\circ}\) is the vertical leg (length 9), and the side opposite \(30^{\circ}\) is the horizontal leg (length \(x\)). Then \(\tan(60^{\circ})=\frac{9}{x}\), so \(x=\frac{9}{\tan(60^{\circ})}=\frac{9}{\sqrt{3}} = 3\sqrt{3}\). But that would mean the side opposite \(30^{\circ}\) is \(3\sqrt{3}\), and opposite \(60^{\circ}\) is 9. But in a \(30 - 60 - 90\) triangle, the side opposite \(30^{\circ}\) is the shortest side. So if 9 is opposite \(60^{\circ}\), then the side opposite \(30^{\circ}\) (the shorter leg) is \(x=\frac{9}{\sqrt{3}} = 3\sqrt{3}\approx5.196\), which is shorter than 9, so that's correct. Then the area is \(\frac{1}{2}\times x\times9=\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\approx23.38\). Wait, but maybe the triangle is labeled differently. Wait, the right angle is between \(x\) and 9, so \(x\) and 9 are the two legs. Then the angle between \(x\) and the hypotenuse is \(60^{\circ}\), so \(\cos(60^{\circ})=\frac{x}{y}\) (hypotenuse \(y\)), and \(\sin(60^{\circ})=\frac{9}{y}\). Since \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}=\frac{9}{y}\), then \(y = \frac{18}{\sqrt{3}}=6\sqrt{3}\). Then \(x = y\cos(60^{\circ})=6\sqrt{3}\times\frac{1}{2}=3\sqrt{3}\). Then area is \(\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\approx23.38\). Alternatively, if we consider the angle of \(60^{\circ}\) between the leg \(x\) and the hypotenuse, then the leg 9 is opposite \(60^{\circ}\), so using the formula for area of a triangle \(A = \frac{1}{2}ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the included angle. Wait, in a right - angled triangle, the included angle between the two legs is \(90^{\circ}\), so \(A=\frac{1}{2}xy\) (legs \(x\) and 9). But we can also use \(A=\frac{1}{2}\times x\times9\). We found \(x = 3\sqrt{3}\), so \(A=\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\approx23.38\). Wait, but let's check with the \(30 - 60 - 90\) ratios. In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\). The side opposite \(30^{\circ}\) is the shortest side (let's call it \(s\)), the side opposite \(60^{\circ}\) is \(s\sqrt{3}\), and the hypotenuse is \(2s\). If the side opposite \(60^{\circ}\) is 9, then \(s\sqrt{3}=9\), so \(s = \frac{9}{\sqrt{3}}=3\sqrt{3}\) (side opposite \(30^{\circ}\)), and the hypotenuse is \(2s = 6\sqrt{3}\). Then the two legs are \(s = 3\sqrt{3}\) (opposite \(30^{\circ}\)) and \(s\sqrt{3}=9\) (opposite \(60^{\circ}\)). Then the area is \(\frac{1}{2}\times3\sqrt{3}\times9=\frac{27\sqrt{3}}{2}\approx23.38\). Yes, that makes sense.

Step2: Calculate the area

The area of a right - angled triangle is given by \(A=\frac{1}{2}\times\text{leg}_1\times\text{leg}_2\). We have one leg \(a = 9\) and the other leg \(b=3\sqrt{3}\) (from the \(30 - 60 - 90\) triangle ratio, where the side opposite \(60^{\circ}\) is \(9\), so the side opposite \(30^{\circ}\) is \(\frac{9}{\sqrt{3}} = 3\sqrt{3}\)).

So \(A=\frac{1}{2}\times9\times3\sqrt{3}=\frac{27\sqrt{3}}{2}\). If we want a decimal approximation, \(\sqrt{3}\approx1.732\), so \(\frac{27\times1.732}{2}=\frac{46.764}{2}=23.382\).

Answer:

The area of the triangle is \(\frac{27\sqrt{3}}{2}\) (or approximately \(23.38\))