Answer:
The distance between the cruise ship and the sailboat is 517.29 feet
Step-by-step explanation:
see the attached figure to better understand the problem
The point A is the top of a cruise ship
The point B is the top of a sailboat
Let
x ---> the distance between the cruise ship and the sailboat (EB=DC)
we know that
In the right triangle ABE
[tex]tan(22^o)=\frac{EA}{EB}[/tex] ---> by TOA (opposite side divided by adjacent side)
we have
[tex]EA=236-27=209\ ft[/tex]
substitute
[tex]tan(22^o)=\frac{209}{EB}[/tex]
solve for EB
[tex]EB=\frac{209}{tan(22^o)}[/tex]
[tex]EB=517.29\ ft[/tex]
therefore
The distance between the cruise ship and the sailboat is 517.29 feet
The distance between the cruise ship and the sailboat is 517.29 feet.
The angle of depression from the top of a cruise ship to the top of a sailboat is 22° . Sitting above water, the cruise ship is 236 feet tall while the sailboat is 27 feet tall.
Let us assume x be the distance between the cruise ship and the sailboat (EB=DC)
So,
Also, we know that
For the right triangle ABE
[tex]tan 22^{\circ} = EA \div EB\\\\So, EA = 236 - 27 = 209\\\\tan 22^{\circ} = 209 \div EB\\\\[/tex]
So, EB = 517.29 feet
learn more about the distance here: https://brainly.com/question/15400849
