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?
○ 5/2 ft
○ 5√3/3 ft
○ 5√3
○ 10

Explanation:

Step1: Use trigonometric ratio

In a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 30^{\circ}\), the adjacent side to \(30^{\circ}\) is the height (\(h\)) and the opposite side is \(5\) ft. Also, \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\).

Step2: Set up the equation

We know that \(\tan30^{\circ}=\frac{5}{h}\) (since \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), opposite = \(5\) ft, adjacent = \(h\)). Substituting \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\), we get \(\frac{1}{\sqrt{3}}=\frac{5}{h}\).

Step3: Solve for \(h\)

Cross - multiply: \(h = 5\sqrt{3}\) (incorrect). Wait, no, we made a mistake. Let's use \(\tan60^{\circ}\). Since \(\tan60^{\circ}=\sqrt{3}\) and \(\tan60^{\circ}=\frac{5}{h}\) (opposite = \(5\), adjacent = \(h\)). Then \(h=\frac{5}{\tan60^{\circ}}\).
Since \(\tan60^{\circ}=\sqrt{3}\), \(h = \frac{5}{\sqrt{3}}\). Rationalize the denominator: \(h=\frac{5\sqrt{3}}{3}\).

Answer:

\(\frac{5\sqrt{3}}{3}\text{ ft}\) (the second option)