Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]D = 11.9874 \ m [/tex]
b
[tex]h = 33.38 \ m [/tex]
Explanation:
From the question we are told that
The length of the anchor above water is L = 4 ft
The length of the anchor inside the water is [tex]d = 36 -4 = 32 \ ft[/tex]
The angle the anchor line makes with the water is [tex]\theta = 22^o[/tex]
Generally from SOHCAHTOA rule
[tex]sin (\theta ) = \frac{D}{d}[/tex]
Here D is the depth of the water
=> [tex]sin(22)= \frac{D}{32}[/tex]
=> [tex]D = 32 * sin(22)[/tex]
=> [tex]D = 11.9874 \ m [/tex]
Generally from SOHCAHTOA rule
[tex]cos (\theta ) = \frac{h}{ d + L}[/tex]
Here h is the horizontal distance
=> [tex]cos (22 ) = \frac{h}{ 4 + 32}[/tex]
=> [tex]cos (22 ) = \frac{h}{ 36}[/tex]
=> [tex]h = 33.38 \ m [/tex]
