# trignometery bearing problem

• Nov 22nd 2006, 04:34 AM
Sohail
trignometery bearing problem
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.:confused:
• Nov 22nd 2006, 06:15 AM
Soroban
Hello, Sohail!

Did you make a sketch?

Quote:

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

Code:

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

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

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

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