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?

Answer 1

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let,  AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

Answer 2

Answer with explanation:

Description of gymnastic mat

Two interior angles of mat which is in the shape of right triangle are 60° and 30°.

Let the base and height of mat are x feet and y feet and Hypotenuse be z feet.

$sin 30=\frac{y}{z}\\\\ \frac{1}{2}=\frac{y}{z}\\\\z=2y$

$Tan30=\frac{y}{x}\\\\x=\sqrt{3}y$

Now, when mat extends 5 feet across the floor,angle of inclination will decrease that is will be less than 30° but the triangle will be a right triangle.

Length of base =(√3 y +5) Feet

Length of Hypotenuse = H feet> z>2 y

The height of gymnastic mat before and after extension will be same.

→→There is another possibility also.

If the total extension of the mat is 5 feet,then

$Base=\sqrt{3}y=5\\\\y=\frac{5}{\sqrt{3}}=\frac{5}{1.732}=2.886$

Length of base of mat =5 feet

Height of mat =2.89 feet(Approx)