The height of the silo is 18 feet
Start by calculating angle ACD using the following tangent trigonometry ratio
[tex]\tan(C) = \frac{AD}{DC}[/tex]
[tex]\tan(C) = \frac{5}{8}[/tex]
[tex]\tan(C) = 0.625[/tex]
Take the arc tan of both sides
[tex]C = \tan^{-1}(0.625)[/tex]
[tex]C = 32^o[/tex]
The measure of angle BCA is then calculated as:
[tex]\angle BCA = 90 - 32[/tex]
[tex]\angle BCA = 58[/tex]
Next, we calculate side length AC using Pythagoras theorem
[tex]AC^2 = AB^2 + BC^2[/tex]
This gives
[tex]AC^2 = 8^2 + 5^2[/tex]
[tex]AC^2 = 89[/tex]
Take the square roots of both sides
[tex]AC = 9.4[/tex]
The height of the silo (length CE) is then calculated using the cosine ratio
[tex]\cos(58) = \frac{9.4}{CE}[/tex]
Make CE the subject
[tex]CE = \frac{9.4}{\cos(58)}[/tex]
[tex]CE = \frac{9.4}{0.53}[/tex]
[tex]CE = 17.7[/tex]
Approximate
[tex]CE = 18[/tex]
Hence, the height of the silo is 18 feet
Read more Pythagoras theorem at:
https://brainly.com/question/654982
