Re: Math problem me and friends are trying to solve

Hello, pierce!

I believe your diagram is wrong . . .

Quote:

A ship leaves port at 10 mph with a heading of N 35° W.

There is a warning buoy located 5 miles directly north of the port.

What is the bearing of the warning buoy **as seen from the ship** after 7.5 hours?

Now I have drawn a diagram and solved that the angle should be 52.7 degrees.

. . Exactly where is that angle?

I just can't figure out what direction.

is it N 52.7 W or would it be N 52.7 E?

How can be the port be to the upper-left or upper-right?

After 10 hours, the ship has traveled 75 miles.

The diagram should look like this:

Code:

` X`

*

:

:

: Y

S * *

: * * :

: * * :

: * * :

* 75 * * B

Z * |

*35d|5

* |

*

P

The port is at $\displaystyle P.$

The ship is at $\displaystyle S.$

The buoy is at $\displaystyle B.$

$\displaystyle \angle BPS = 35^o,\;SP = 75$

Law of Cosines:

. . $\displaystyle SB^2 \;=\;5^2 + 75^2 - 2(5)(75)\cos35^o \;=\;5035.635967 $

. . .$\displaystyle SB \;=\;70.96221506 \;\approx\;71\text{ miles}$

Law of Sines:

. . $\displaystyle \frac{\sin(SBP)}{75} \:=\:\frac{\sin(SPB)}{71} \quad\Rightarrow\quad \sin(SBP) \:=\:\frac{75\sin(SPB)}{71} \:=\:0.605890602$

. . . . $\displaystyle \angle SBP \;=\;37.29295617 \;=\;37.3^o\text{ or }142.7^o$

We see that $\displaystyle \angle SBP$ is obtuse: .$\displaystyle \angle SBP \,=\,142.7^o$

Then: .$\displaystyle \angle XSB \,=\,142.7^o \quad\Rightarrow\quad \angle BSZ \,=\,37.3^o$

Therefore, the heading of the buoy from the ship is: .$\displaystyle S\,37.3^o\,E$