Ashark is swimming 28 feet below sea level. if the angle of
depression from a boat on the water to the shark is 13º, what is the
horizontal distance between the boat and the shark?

The horizontal distance between boat and the shark is 121.28 feet

Step-by-step explanation:

Given as :

The Distance of the shark swimming below sea level = AB = 28 feet

The angle of depression from a boat on water to shark = 13°

Let The horizontal distance between shark and boat = OA =  x

Now, Tan angle =

or, Tan 13° =  

Or, Tan 13° =  

Or, 0.2308 =  

Or, x =  

∴ x = 121.28 feet

So The horizontal distance = x = 121.28 feet

Hence The horizontal distance between boat and the shark is 121.28 feet Answer

121.28 feet

Step-by-step explanation:

See the diagram attached.

Let B is the position of the boat and BA = 28 feet is the height of the water surface from the level of the shark.

If the position of the shark is at S then angle of depression from boat to the sherk i.e. ∠ OBS = 13° and we have ∠ BSA = ∠ OBS = 13° {Alternate angles}

Now, using trigonometry in right triangle Δ ABS we can write,  

⇒ feet.

Therefore, the horizontal distance between the boat and the shark is AS = 121.28 feet, (Answer)

could you reply to this with the question so i can ?

answer: 230 miles ( vote ! )

step-by-step explanation:

180.39 minus 17.99 equals 162.4

then divide 162.4 by .80 to get 230