Areconnaissance airplane p, flying at 19,000 feet above a point r on the surface of the water, spots a submarine s at an angle of depression of β = 21° and a tanker t at an angle of depression of α = 33°, as shown in the figure. in addition, ∠spt is found to be γ = 119°. approximate the distance between the submarine and the tanker.

The solution for the problem is:

Tan 33(degrees) = opposite/adjacent 

Starting off with the submarine, find how far it is from point R. 

Tan 21 = 19,000/x 

or 

19,000/Tan 21= x 

x = 49496.69 feet the submarine is from point R. 

Now, find how far the tanker is from point R. 

Tan 33= 19,000/x 

or 

19,000/Tan 33 = x 

x = 29257.43 feet the tanker is from point R. 

Now subtract and the difference would be the answer

= 49496.69 - 29257.43

The tanker and sub are 20239.26  feet away from each other. 

step-by-step explanation:

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