trignometery bearing problem

**Sohail** · November 22nd 2006, 04:34 AM

A point P is 12 km due north of another point Q. the bearing of a lighthouse,R, from P is 135 degress and, from Q, it is 120 degress. calculate the distance PR.

**Soroban** · November 22nd 2006, 06:15 AM

Hello, Sohail!

Did you make a sketch?

A point $P$ is 12 km due north of another point $Q$ .
The bearing of a lighthouse $R$ from $P$ is 135°, and from $Q$ it is 120°.
Calculate the distance $PR$ .

Code:

      |
    P * 135°
      | *
      |45°*
 r=12 |     *
      |       *
      |         *
    Q * 120°      *
         *          * q
            *         *
               *        *
                  *       *
                     *      *
                        *     *
                           *    *
                              *15°*
                                 *  * 
                                      * R

In $\Delta PQR$ , we have: . $\angle P = 45^o,\;\angle Q = 120^o$ , then $\angle R = 15^o$
. . also: .side $r = PQ = 12$ and we want side $q = PR.$

Law of Sines: . $\frac{q}{\sin Q} \,= \,\frac{r}{\sin R}\quad\Rightarrow\quad q \:=\:\frac{r\sin Q}{\sin R}$

Therefore, we have: . $q\:=\:PQ\:=\:\frac{12\cdot\sin120^o}{\sin15^o} \:\approx\:40.15$ km.

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