1. ## BEARING/trig problem

From a point P, the ships R, V are observed bearings 325degreesT and 47degreesT respectively, whilst from a point Q 10km due north of P the ships R, V bear 310degreesT and 79degreesT respectively.
i) Show that PQ = 10sin50/sin15 and obtain similar expression for PV

dont realli get how to draw the diagram at all , doesnt make sense to me =/// !!

any help is GREATLY appreciated !!!
thankyou !

2. Originally Posted by iiharthero
From a point P, the ships R, V are observed bearings 325degreesT and 47degreesT respectively, whilst from a point Q 10km due north of P the ships R, V bear 310degreesT and 79degreesT respectively.
i) Show that PQ = 10sin50/sin15 and obtain similar expression for PV

dont realli get how to draw the diagram at all , doesnt make sense to me =/// !!

any help is GREATLY appreciated !!!
thankyou !
All the angles are measured with respect to the North direction, clockwise!

See attachment.

3. Hello, iiharthero!

From a point $P$, the ships $R, V$ are observed at 325° and 47°, resp,
while from a point $Q$ 10km due north of $P$, the ships $R, V$ bear 310° and 79°, resp.

i) Show that $PR \:=\: \frac{10\sin50^o}{\sin15^o}$ and obtain similar expression for $PV$.

This is the diagram . . .

Code:
                                            o V
* *
|              *  *
R o               |           *   *
* *           |        *    *
*   *   50° | 79° *     *
*     *   |  *      *
*       oQ      *
*     |     *
*35°|47°*
* | *
o
P
where $PQ = 10$

Consider $\Delta PQR$ and all its angles.
Code:
    R o
* *
*15°*           |
*     *   50° |
*       *   |
*    130° o Q
*       |
*     |10
*35°|
* |
o
P

Law of Sines: . $\frac{PR}{\sin130^o} \:=\:\frac{10}{\sin15^o} \quad\Rightarrow\quad PR \:=\:\frac{10\sin130^o}{\sin15^o}$

Since $\sin130^o = \sin50^o$, we have: . $PR \:=\:\frac{10\sin50^o}{\sin15^o}$