Question

From the side view, a gymnastics mat forms a right

Answer 1

Answer:

its b

Step-by-step explanation:

Answer 2

Answer:

$h=\frac{5\sqrt{3} }{3} \,\,m$    

Step-by-step explanation:  

Let BC represents 5 feet across the floor, and AB represents the height.

Consider the figure in the attached file.

We know that,

$\text{tan}\theta =\frac{\text{Perpendicular}}{\text{Base}}$

$\implies\text{tan}\,C\ =\frac{\text{AB}}{\text{BC}}$

Here, $\angle C=30^{\circ},\text{AB}=h\:\text{meters} , \text{and}\:\text{BC}=5\:\text{meters}$

So, we have

$\text{tan\,30}^{\circ}=\frac{h}{5}$

$\implies \frac{1}{\sqrt{3}}=\frac{h}{5}\:\:\left [ \because \text{tan\,30}^{\circ}=\frac{1}{\sqrt{3}} \right ]$

$\implies h=\frac{5}{\sqrt{3} }$

$\implies h=\frac{5\times \sqrt{3}}{\sqrt{3}\times \sqrt{3} }$

$\implies h=\frac{5\sqrt{3}}{3 } \,m$

Hence, the mat is $\frac{5\sqrt{3}}{3 } \,m$ high of the ground.