Thursday

October 23, 2014

October 23, 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

**Answer this Question**

**Related Questions**

Trigonometry - A person on a ship sailing due south at the rate of 15 miles an ...

Algebra - Sorry for asking another question, but I don't know how to set this ...

Mathamatics - A ship is due south of a lighthouse. It sails on a bearing of 72* ...

Algebra 3 and trig - An ocean liner is 177 miles due west of lighthouse A. ...

maths - A ship sailing on a course bearing 036 degrees is 5500 metres due south...

CALCULUS!!!!! - At 9am, ship B was 65 miles due east of another ship, A. Ship B ...

calculus - At 9am, ship B was 65 miles due east of another ship, A. Ship B was ...

calculus - Optimization At 1:00 PM ship A is 30 miles due south of ship B and is...

Calculus - one ship (A) is sailing due south at 16 miles per hour and a second ...

Calculus - One ship is 20 miles due North of another ship, and is sailing South ...