Question

from the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. the gymnastics mat extends 5 feet across the floor. how high is the mat off the ground?
options:

• \\(\frac{5}{2}\\) ft
• \\(\frac{5\sqrt{3}}{3}\\) ft
• \\(5\sqrt{3}\\)
• 10

Explanation:

Step1: Identify the trigonometric relationship

We have a right triangle with one angle \(30^\circ\), the adjacent side to \(30^\circ\) is the height (\(h\)) and the opposite side is \(5\) feet. Wait, no, actually, the angle of \(30^\circ\) has the adjacent side as the base (5 ft) and the opposite side as the height? Wait, no, let's check the triangle. The right angle, one angle \(60^\circ\), one \(30^\circ\). The base (across the floor) is 5 ft, which is adjacent to the \(30^\circ\) angle? Wait, no, the tangent of an angle in a right triangle is \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). Let's define: let the angle be \(30^\circ\), the adjacent side to \(30^\circ\) is the height? No, wait, the base is 5 ft, which is adjacent to the \(30^\circ\) angle? Wait, no, the triangle: right angle, one angle \(60^\circ\) (at the top), one \(30^\circ\) (at the bottom right). So the base (horizontal side) is 5 ft, which is adjacent to the \(30^\circ\) angle, and the height (vertical side) is opposite to the \(30^\circ\) angle? Wait, no, \(\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{height}}{5}\)? Wait, no, \(\tan(60^\circ)=\frac{5}{\text{height}}\)? Wait, let's correct. Let's denote the height as \(h\), the base (horizontal) as 5 ft. The angle at the bottom is \(30^\circ\), so the angle of \(30^\circ\) has adjacent side \(h\) and opposite side 5 ft? No, that's not right. Wait, 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. Wait, the side across the floor (base) is 5 ft, which is adjacent to the \(60^\circ\) angle? Wait, maybe better to use \(\tan(60^\circ)=\frac{5}{h}\), so \(h = \frac{5}{\tan(60^\circ)}\). Since \(\tan(60^\circ)=\sqrt{3}\), so \(h=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\)? Wait, no, wait. Wait, the angle of \(30^\circ\): the side opposite \(30^\circ\) is the height, and the side adjacent is 5 ft? No, that can't be. Wait, let's draw the triangle: right angle at the bottom left, vertical side (height) on the left, horizontal side (base) at the bottom (5 ft), hypotenuse connecting top left to bottom right. The angle at the top left is \(60^\circ\), so the angle at the bottom right is \(30^\circ\). So for the angle \(30^\circ\) (at bottom right), the adjacent side is the base (5 ft), and the opposite side is the height (h). So \(\tan(30^\circ)=\frac{h}{5}\)? No, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), so \(\tan(30^\circ)=\frac{h}{5}\), so \(h = 5\tan(30^\circ)\). But \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), so \(h = 5\times\frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\). Wait, but let's check the 30-60-90 triangle ratios. In a 30-60-90 triangle, the sides are in the ratio \(x : x\sqrt{3} : 2x\), where \(x\) is the side opposite \(30^\circ\), \(x\sqrt{3}\) opposite \(60^\circ\), and \(2x\) hypotenuse. Wait, if the side opposite \(60^\circ\) is 5 ft (the base), then the side opposite \(30^\circ\) (height) is \(\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\)? Wait, no, the side opposite \(60^\circ\) is the base (5 ft), so in the ratio, the side opposite \(60^\circ\) is \(x\sqrt{3}\), so \(x\sqrt{3}=5\), so \(x=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\), and the side opposite \(30^\circ\) is \(x\), which is the height. So that's correct. So the height is \(\frac{5\sqrt{3}}{3}\) ft.

Step2: Calculate the height

We know that in a right triangle, \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{5}{h}\), where \(h\) is the height. Since \(\tan(60^\circ)=\sqrt{3}\), we have \(\sqrt{3}=\frac{5}{h}\), so solving for \(h\) gives \(h = \frac{5}{\sqrt{3}}\). Rationalizing the denominator, we multiply numerator and denominator by \(\sqrt{3}\): \(h=\frac{5\sqrt{3}}{3}\) ft.

Answer:

\(\frac{5\sqrt{3}}{3}\) ft (corresponding to the option \(\boldsymbol{\frac{5\sqrt{3}}{3}}\) ft)