right
[tex]\begin{gathered} AI)\text{ 400 ft} \\ MI)412.31\text{ f} \\ \text{angle = 76} \end{gathered}[/tex]
Explanation
Step 1
AI?
we have a rigth triangle
then
let
[tex]\begin{gathered} AB=side1 \\ AI=side\text{ 2} \\ IB=\text{ hypotenuse} \end{gathered}[/tex]
we can use the pythagorean Thoerem to find the missing vale
so
[tex]\begin{gathered} (AB)^2+(AI)^2=(BI)^2 \\ \text{replace} \\ 300^2+(AI)^2=500^2 \\ so \\ (AI)^2=500^2-300^2 \\ AI=\sqrt[]{500^2-300^2}=\sqrt[]{160000}=400 \\ AI=400 \end{gathered}[/tex]
Step 2
MI?
let
[tex]\begin{gathered} \text{angle}=x \\ \text{opposite side=100 m} \\ \text{adjacent side=400 m} \end{gathered}[/tex]
so, we need a function that relates those 3 values
[tex]\tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}[/tex]
replace
[tex]\begin{gathered} \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \tan x=\frac{400}{100} \\ \tan x=4 \\ \text{hence} \\ x=\tan ^{-1}(4) \\ x=75.96 \\ \text{rounded} \\ x=76\text{ \degree} \end{gathered}[/tex]
As 76 is greater than 68, the zipline cable compliance with these regulations.
Also, the hypotenuse (zipline ) is
[tex]\begin{gathered} (MI)^2=(AI)^2+(AM)^2 \\ \text{replace} \\ (MI)^2=(400)^2+(100)^2 \\ (MI)^2=170000 \\ MI=\sqrt[]{17000} \\ MI=412.31\text{ ft} \end{gathered}[/tex]
I hope this helps you
