Hello, iiharthero!

From a point $\displaystyle P$, the ships $\displaystyle R, V$ are observed at 325° and 47°, resp,

while from a point $\displaystyle Q$ 10km due north of $\displaystyle P$, the ships $\displaystyle R, V$ bear 310° and 79°, resp.

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

This is the diagram . . .

Code:

o V
* *
| * *
R o | * *
* * | * *
* * 50° | 79° * *
* * | * *
* oQ *
* | *
*35°|47°*
* | *
o
P

where $\displaystyle PQ = 10$

Consider $\displaystyle \Delta PQR$ and all its angles. Code:

R o
* *
*15°* |
* * 50° |
* * |
* 130° o Q
* |
* |10
*35°|
* |
o
P

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

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