Posted by **AwesomeGuy** on Monday, February 4, 2013 at 10:13pm.

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?

- Trigonometry -
**Reiny**, Monday, February 4, 2013 at 11:01pm
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.

- Trigonometry -
**AwesomeGuy**, Tuesday, February 5, 2013 at 1:56am
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!

- Trigonometry -
**AwesomeGuy**, Tuesday, February 5, 2013 at 2:04am
Since I have also forgot to label the conversion for the solution, it is ultimately measured in FEET.

- Trigonometry -
**Meh**, Thursday, March 12, 2015 at 11:49am
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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