bearing

• Sep 6th 2010, 04:31 AM
aeroflix
bearing
Attachment 18840

kindly answer the bearing... i got the angle at forest island but i dont know how to get the bearing in from forst island to home port..

the bearing should be South East, but what degrees
• Sep 6th 2010, 08:49 AM
Soroban
Hello, aeroflix!

Quote:

A fisherman leaves his homeport and heads in the direction $N\,70^o\,W$.
He sails 30 miles and reaches Egg Island.

The next day he sails $N\,10^o\,E$ for 50 miles to Forest Island.

(b) Find the bearing from Forest Island back to his homeport.

In my diagram, I omitted the distances (they wouldn't fit).
I hope you read it.

Code:

            F             ♥       Q    *:       :  * :*       :10*  :       : *  : *  P       :*    S    :     E ♥100    *  :       : *        :       :70 *    * :       :    *  70:       R      *  *:                 * :                   ♥ H

The homeport is $H.$

He sails 30 miles to $E.$
. . $\angle PHE = \angle HER = 70^o,\;HE = 30$

Then he sails 50 miles to $F.$
. . $\angle QEF = 10^o \quad\Rightarrow\quad \angle FEH = 100^o,\;\; EF = 50$

I assume that part (a) asked for the distance from $F$ to $H.$

. . $FH^2 \;=\;30^2 + 50^2 - 2(30)(50)\cos100^o \;=\;3920.944533$

Hence: . $FH \;=\;62.61744592 \;\approx\;62.62$

Now we determine $\angle FHE.$

. . $\cos(\angle FHE) \;=\;\dfrac{30^2 + 62.62^2 - 50^2}{2(30)(62.62)} \;=\; 0.61781763$

Hence: . $\angle FHE \;=\;52.84305918^o \;\approx\;51.8^o$

Then:. . $\angle PHF \;=\;70^o - 51.8^o \;=\;18.2^o$

Note that: . $\angle SFH = \angle PHF$

Therefore, the bearing is: . $S\,18.2^o\,E$

• Sep 6th 2010, 12:41 PM
aeroflix
thanks soroban.. now i realize that there are some same angles :)