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December 19, 2014

December 19, 2014

Posted by **Amber** on Saturday, June 23, 2012 at 10:31am.

Please Show The Solution So That I Can Study the Problem. Thanks In Advance :)

- Trigonometry -
**bobpursley**, Saturday, June 23, 2012 at 10:54amdraw the picture as a at 3pm and at 5 pm. Label the distance the ship from the ship to the lighthouse as d.

Note d^2=original distance^2+(15t)^2

where t=0 at 3pm, t=2 at 5pm, etc.

Now, at t=2, using law of sines...

original distance/sin(90-52)=d/sin90

so you know then d=originaldistance/sin38 at 5PM

also, from the law of cosines..

originaldistance^2=d^2+(15t)^2-2d15tcos38 where t=2

now, you can solve for all.

- Trigonometry -
**irish**, Thursday, November 21, 2013 at 6:10am24.36

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