Trigonometry
posted by AwesomeGuy .
An observer in a lighthouse 350 feet above sea level observes two ships directly offshore. The angles of depression to the ships are 4° and 6.5°. How far apart are the ships?

My diagram has the lighthouse as PQ with P at the top and PQ = 350
My ships are at A and B, angle at A = 4° and the angle at B = 6.5°
In the right angled triangle BQP
sin 6.5 = 350/BP
BP = 350/sin 6.5 = .....
now look at triangle ABP
we just found BP
and angle ABP = 173.5°
thus angle APB = 180  4  173.5 = 2.5°
by the sine law:
AB/sin 2.5 = BP/sin 4°
I will let you finish this, let me know what you got. 
BP= 3091.79
AB/sin 2.5° = 3091.79/sin 4°
3091.785015 sin 2.5° = 134.8617682
134.8617682/sin 4° = 1933.33
Awesome! Thanks for the help! 
Since I have also forgot to label the conversion for the solution, it is ultimately measured in FEET.

Draw two triangles upside down.
For the first triangle, the angle between the base and the hypotenuse is 4 degrees, and the vertical is 350.
The second triangle, the angle between the base and the hypotenuse will be 6.5 degrees and the vertical is also 350.
Now to find the distance :
1st triangle: 350/tan 4
2nd triangle 350/tan 6.5
now subtract both and you will get 1933.3 feet.
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