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,
[tex]\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}[/tex]
Similarly, in Δ ACD,
[tex]\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[/tex]
Hence, the mat is 4.33 ft high off the ground.
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.
[tex]sin 30=\frac{y}{z}\\\\ \frac{1}{2}=\frac{y}{z}\\\\z=2y[/tex]
[tex]Tan30=\frac{y}{x}\\\\x=\sqrt{3}y[/tex]
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
[tex]Base=\sqrt{3}y=5\\\\y=\frac{5}{\sqrt{3}}=\frac{5}{1.732}=2.886[/tex]
Length of base of mat =5 feet
Height of mat =2.89 feet(Approx)
