Question

find the value of x. diagram: right triangle (right angle, vertical leg ( x ), horizontal leg ( 28 ), ( 30^circ ) angle at bottom). options: a ( \frac{56}{3}sqrt{3} ), b ( 28sqrt{3} ), c ( \frac{28}{3}sqrt{3} ), d ( 56 ). right: three reference triangles (30-60-90, 45-45-90, general right triangle).

Explanation:

Step1: Identify the triangle type

This is a 30 - 60 - 90 right 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 side (let's call it \(n\)), the side opposite \(60^{\circ}\) (adjacent to \(30^{\circ}\)) is \(n\sqrt{3}\), and the hypotenuse is \(2n\). Here, the side adjacent to \(30^{\circ}\) is 28, and the side opposite \(30^{\circ}\) is \(x\)? Wait, no. Wait, in the given triangle, the right angle, \(30^{\circ}\) angle, so the side adjacent to \(30^{\circ}\) is 28, and the side opposite \(30^{\circ}\) is \(x\)? Wait, no, let's correct. In a right - triangle with angle \(30^{\circ}\), \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{28}\). Wait, no, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), here \(\theta = 30^{\circ}\), opposite side is \(x\), adjacent side is 28. But also, in 30 - 60 - 90 triangle, if the adjacent side to \(30^{\circ}\) (which is the longer leg, opposite \(60^{\circ}\)) is \(n\sqrt{3}\), and the shorter leg (opposite \(30^{\circ}\)) is \(n\). Wait, the adjacent side to \(30^{\circ}\) is 28, which is the longer leg (opposite \(60^{\circ}\)), so \(n\sqrt{3}=28\), then \(n=\frac{28}{\sqrt{3}}\), but the shorter leg (opposite \(30^{\circ}\)) is \(n\)? No, wait, no. Wait, the side opposite \(30^{\circ}\) is the shorter leg, side opposite \(60^{\circ}\) is the longer leg. So if the longer leg (opposite \(60^{\circ}\)) is 28, then \(\tan(60^{\circ})=\frac{x}{28}\)? No, wait, the angle is \(30^{\circ}\), so the angle at the base is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is \(x\), and the side adjacent to \(30^{\circ}\) is 28. So \(\tan(30^{\circ})=\frac{x}{28}\), but \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(x = \frac{28}{\sqrt{3}}\), but that's not one of the options. Wait, maybe I mixed up. Wait, in the 30 - 60 - 90 triangle, the ratio of sides is: shorter leg (opposite \(30^{\circ}\)) : longer leg (opposite \(60^{\circ}\)) : hypotenuse = \(n:n\sqrt{3}:2n\). So if the longer leg (opposite \(60^{\circ}\)) is 28, then \(n\sqrt{3}=28\), so \(n=\frac{28}{\sqrt{3}}\), but the shorter leg is \(n\), and the hypotenuse is \(2n\). Wait, but the side \(x\) is opposite \(30^{\circ}\)? No, wait the triangle has a right angle, \(30^{\circ}\) angle, so the sides: let's label the triangle. Let the right angle be at the bottom left, so the vertical side is \(x\) (opposite \(30^{\circ}\)), horizontal side is 28 (adjacent to \(30^{\circ}\)), and hypotenuse is the other side. So \(\tan(30^{\circ})=\frac{x}{28}\), so \(x = 28\times\tan(30^{\circ})\). But \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\), so \(x=\frac{28\sqrt{3}}{3}\)? No, wait that's option C? Wait no, wait maybe I got the angle wrong. Wait, maybe the angle is \(60^{\circ}\) at the base. Wait, the triangle has a right angle, and one angle is \(30^{\circ}\), so the other angle is \(60^{\circ}\). So if the horizontal side is 28 (adjacent to \(30^{\circ}\)), then the vertical side \(x\) is opposite \(30^{\circ}\), and the hypotenuse is opposite \(90^{\circ}\). Alternatively, \(\cot(30^{\circ})=\frac{28}{x}\), since \(\cot\theta=\frac{\text{adjacent}}{\text{opposite}}\). \(\cot(30^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{28}{x}\), then \(x=\frac{28}{\sqrt{3}}=\frac{28\sqrt{3}}{3}\)? No, that's not matching. Wait, no, the 30 - 60 - 90 triangle ratios: if the shorter leg (opposite \(30^{\circ}\)) is \(n\), then the longer leg (opposite \(60^{\circ}\)) is \(n\sqrt{3}\), and hypotenuse is \(2n\). So in the given triangle, the horizontal side is 28, which is the longer leg (opposite \(60^{\circ}\)), so \(n\sqrt{3}=28\), so \(n = \frac{28}{\sqrt{3}}=\frac{28\sqrt{3}}{3}\). But the vertical side \(x\) is the shorter leg (opposite \(30^{\circ}\)), so \(x = n=\frac{28\sqrt{3}}{3}\)? No, that's option C? Wait, no, wait the options are A: \(\frac{56}{3}\sqrt{3}\), B: \(28\sqrt{3}\), C: \(\frac{28}{3}\sqrt{3}\), D: 56. Wait, maybe I made a mistake. Wait, let's use tangent of \(60^{\circ}\). The angle at the base is \(30^{\circ}\), so the angle at the top (other than right angle) is \(60^{\circ}\). So \(\tan(60^{\circ})=\frac{x}{28}\), since \(\tan(60^{\circ})=\sqrt{3}\), so \(x = 28\sqrt{3}\), which is option B. Ah! I see, I mixed up the angle. The side \(x\) is opposite \(60^{\circ}\), and the side 28 is adjacent to \(60^{\circ}\) (since the angle at the base is \(30^{\circ}\), so the angle at the top is \(60^{\circ}\)). So \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{28}\), and \(\tan(60^{\circ})=\sqrt{3}\), so \(x = 28\times\sqrt{3}\).

Step2: Calculate \(x\)

Using the tangent function for the \(60^{\circ}\) angle in the right - triangle:
\(\tan(60^{\circ})=\frac{x}{28}\)
We know that \(\tan(60^{\circ})=\sqrt{3}\), so:
\(\sqrt{3}=\frac{x}{28}\)
Multiply both sides by 28:
\(x = 28\sqrt{3}\)

Answer:

B. \(28\sqrt{3}\)