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?
height
5 ft
30°
60°
5 ft
\\( \frac { 5 } { 2 } \mathrm { ft } \\)
\\( \frac { 5 \sqrt { 3 } } { 3 } \mathrm { ft } \\)
\\( 5 \sqrt { 3 } \\)
10

Explanation:

Step1: Use trigonometric ratio

In a right - triangle, we can use the tangent function. The formula for tangent is \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 30^{\circ}\), the opposite side is the height \(h\) (what we want to find) and the adjacent side is \(5\) ft. So, \(\tan30^{\circ}=\frac{h}{5}\).

Step2: Recall the value of \(\tan30^{\circ}\)

We know that \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\). Substituting this into the equation \(\frac{1}{\sqrt{3}}=\frac{h}{5}\).

Step3: Solve for \(h\)

Cross - multiply to get \(h = \frac{5}{\sqrt{3}}\). Rationalize the denominator: \(h=\frac{5\sqrt{3}}{3}\) (by multiplying numerator and denominator by \(\sqrt{3}\)).

Answer:

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