Hello, ferken!
This is a 3D trig problem, but we can work in two dimensions at a time.
Jason is on a 50m high bridge and sees two boats anchored below.
From his position, boat A has a bearing of 230° and boat B has a bearing of 120°.
Jason estimates the angles of depression to be 38° for boat A and 35° for Boat B.
How far apart are the boats (nearest meter)? Jason looks at boat A. Code:
W      * J
38° * 
* 
*  50
* 
* 38° 
A *      * C
Jason is at $\displaystyle J.$
The bridge is: $\displaystyle JC = 50$
$\displaystyle \angle WJA = 38^o = \angle JAC$
In right triangle $\displaystyle JCA\!:\;\;\tan38^o \:=\:\frac{50}{AC} \quad\Rightarrow\quad AC \:=\:\frac{50}{\tan38^o} \:\approx\:64.0$
Jason looks at boat B. Code:
J *     E
 * 35°
 *
50  *
 *
 35° *
C *      * B
Jason is at $\displaystyle J.$
The bridge is: $\displaystyle JC = 50$
$\displaystyle \angle EJB = 35^o = \angle JBC$
In right triangle $\displaystyle JCB\!:\;\;\tan35^o \:=\:\frac{50}{BC} \quad\Rightarrow\quad BC \:=\:\frac{50}{\tan35^o} \:\approx\:71.4$
Looking down from the sky, this is the diagram: Code:
N



C* 120°
64 * *
* 110° * 71.4
A* *
* *
* *
*B
Law of Cosines: .$\displaystyle AB^2 \;=\;64^2 + 71.4^2  2(64)(71.4)\cos110^o \;=\;12319.75049$
Therefore: .$\displaystyle AB \;=\;110.9943715 \;\approx\;111\text{ m}$