1. ## 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.

2. Hello, Sohail!

Did you make a sketch?

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.

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# a point p is 40 km from q on a bearing 061 calculate the distance that p is

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