# Thread: Navigation: Using law of cosines

1. ## Navigation: Using law of cosines

A fisherman leaves his home port and heads in the direction N 70 degree W. He travels 30 mi and reaches Egg Island. The next day he sails N 10 degree E for 50 mi, reaching Forrest Island.

a) Find the distance between the fisherman's home port and Forrest Island.

b) Find the bearing from Forrest Island back to his home port.

I do not understand part b.

2. You need to draw a picture to understand this problem fully

my solution for $d$the distance between forrest island and the fisherman's home is $d^2 = 50^2+30^2-2\times 50 \times 30 \times \cos(140)$

ths gives $d \approx 63$

3. Let's do it this way by traversing from home to Egg Island, then to Forrest Island, then back home. I am used to using azimuths. Azimuths are easy because they just turn right around the clock. Bearings are done in quadrants.

From home to Egg Island, the coordinates of Egg Island are given by:

$(30sin(290), \;\ 30cos(290))\approx (-28.19, \;\ 10.26)$

290 is the azimuth turned clockwise. 360-70=290.

So far, so good?.

Now, from there to Forrest Island. The coordinates of Forrest Island are:

$(30sin(290)+50sin(10), \;\ 30cos(290)+50cos(10))\approx (-19.50, \;\ 59.50)$

See?. We find the coordinates by adding to the previous ones at Egg Island.

Now, the distance back home can be found by using Pythagoras:

$\sqrt{(30sin(290)+50sin(10))^{2}+(30cos(290)+50cos (10))^{2}}\approx 62.617$

The bearing from home to Forrest Island is just

$tan^{-1}\left(\frac{30sin(290)+50sin(10)}{30cos(290)+50c os(10)}\right)\approx -18.15^{o} or \;\ 341.8476^{o}$

The answer will be negative. This means it is to the left of the y-axis or north. Add to 360 to find the azimuth.

By subtracting 180, we find the bearing from Forrest to home the other direction: $161.847^{o}$

4. Hello, musicbeat!

pickslides is absolutely correct!

A fisherman leaves his home port and heads in the direction N 70° W.
He travels 30 mi and reaches Egg Island.
he next day he sails N 10° E for 50 mi, reaching Forrest Island.

a) Find the distance between the fisherman's home port and Forrest Island.

b) Find the bearing from Forrest Island back to his home port.
Code:
                  F
N         o
|       *
| 10° *         Q
|   * 50        |
| *             |
E o 100°          |
|   *   30      |
|70°    *    70°|
|           *   |
|               o
S               P

The fisherman's port is $P.$
He sails 30 miles to Egg Island $(E)\!:\;\angle QPE = 70^o,\;PE = 30$
Then he sails 50 miles to Forrest Island $(F)\!:\;\angle NEF = 10^o,\;EF = 50$
Draw line $FP.$

Since $NS \parallel QP\!:\;\;\angle SEP = \angle QPE = 70^o$
. . Hence: . $\angle FEP \,=\,100^o$

Using the Law of Cosines on $\Delta FEP$, we have:

. $d^2 \:=\:F\!P^2 \:=\:50^2 + 30^2 - 2(50)(30)\cos100^o \;=\;3920.944533$

Therefore: . $d \;=\;62.61744592 \;\approx\;63\text{ miles}$

Again, using the Law of Cosines:

. . $\cos(\angle FPE) \:=\:\frac{30^2 + 63^2 - 50^2}{2(30)(63)} \;=\;0.626719577$

Hence: . $\angle FPE \;=\;51.13148859^o \;\approx\;51^o$

Then: . $\angle FPQ \;=\;70^o - 51^o \;=\;19^o$

The heading from $F$ to $P$ is: . $S\,19^o\,E$

5. So am I. Just done a different way by using a traverse and azimuths.