bearing

**aeroflix** · September 6th 2010, 03:31 AM

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

**Soroban** · September 6th 2010, 07:49 AM

Hello, aeroflix!

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: . $S\,18.2^o\,E$

**aeroflix** · September 6th 2010, 11:41 AM

thanks soroban.. now i realize that there are some same angles

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