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

A person on a ship sailing due south at the rate of 15 miles an hour observes a lighthouse due west at 3p.m. At 5p.m. the lighthouse is 52degrees west of north. How far from the lighthouse was the ship at a)3p.m.? b)5p.m.? c)4p.m.?

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

  • Trigonometry - , Saturday, June 23, 2012 at 10:54am

    draw 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 - , Thursday, November 21, 2013 at 6:10am


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