# Math Help - trig word problem, bearings

1. ## trig word problem, bearings

This is the problem: A ship is sailing due north. At a certain point the bearing of a lighthouse is N 38 E and the distance is 13.5. After a while, the captain notices that the bearing of the lighthouse is now S 59.4 . How far did the ship travel between the two observations of the lighthouse. distance=?
Here is what I did: at A the ship tskes a bearing of the lighthouse at C and finds angle A=38 and b=AC=13.5 miles. Later at point B, they find that angle B=59.4. So, angleC= 180-38-59.4=82.6. Next we find c, I used the law of sines, 13.5sin82.6/sin59.4= 15.6 miles.

I checked and my answer is wrong, so could someone please show me what I did wrong.
Thank you!
Keith Stevens

2. Originally Posted by kcsteven
This is the problem: A ship is sailing due north. At a certain point the bearing of a lighthouse is N 38 E and the distance is 13.5. After a while, the captain notices that the bearing of the lighthouse is now S 59.4 . How far did the ship travel between the two observations of the lighthouse. distance=?
Here is what I did: at A the ship tskes a bearing of the lighthouse at C and finds angle A=38 and b=AC=13.5 miles. Later at point B, they find that angle B=59.4. So, angleC= 180-38-59.4=82.6. Next we find c, I used the law of sines, 13.5sin82.6/sin59.4= 15.6 miles.

I checked and my answer is wrong, so could someone please show me what I did wrong.
Thank you!
Keith Stevens
Looks OK to me: 15.553511. Checks out on a scale diagram to the limits
of accuracy of the drawing.

RonL

3. Hello, Keith!

I see absolutely nothing wrong with your work . . .

A ship is sailing due north. .At a certain point
the bearing of a lighthouse is $N\,38^o\,E$ and the distance is $13.5$ miles.
After a while, the captain notices that the bearing of the lighthouse is now $S\,59.4^o\,E$
How far did the ship travel between the two observations of the lighthouse?

Here is what I did:
At $A$ the ship takes a bearing of the lighthouse at $C$ and finds $\angle A = 38^o,\;b = AC = 13.5$ miles.

Later at point $B$, they find that $\angle B = 59.4^o.$
So: . $\angle C\:= \:180 - 38 - 59.4 \:= \:82.6^o$

Next we find $c$,
I used the Law of Sines: . $c \:=\:\frac{13.5\sin82.6^o}{\sin59.4^o} \:= \:15.6$ miles.

I checked and my answer is wrong.

Either: [1] You copied one of the numbers incorrectly