Answer:
[tex]Length\hspace{3}of\hspace{3}the\hspace{3}beam=y=21.3ft[/tex]
Step-by-step explanation:
Look the picture I attached you. As you can see the beam against the wall form a right triangle. The trigonometry functions on a right triangle are:
[tex]sin(\theta)=\frac{opposite}{hypotenuse}\hspace{10}csc(\theta)=\frac{hypotenuse}{opposite}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\hspace{10}sec(\theta)=\frac{hypotenuse}{adjacent} \\\\tan(\theta)=\frac{opposite}{adjacent} \hspace{20}cot(\theta)=\frac{adjacent}{opposite}[/tex]
The problem give us the following data:
[tex]y=9sec(\theta)[/tex]
Using the previous information about the trigonometry functions on a right triangle and the data provided by the problem you can conclude:
[tex]y=Hypotenuse\\Adjacent=9\\\theta=65^{\circ}[/tex]
Therefore:
[tex]y=9sec(65^{\circ})=9*(2.366201583)=21.29581425\approx21.3ft[/tex]