Answer
The boats are $\approx 1412.0$ meters apart.
Work Step by Step
There are two main triangles present in the diagram (Figure 1, see attachment):
$\triangle ABC$
$\triangle ACD$
$AB =$ Height of the plane
$CD =$ Distance between the boats
1. Find all the angles in $\triangle ABC$
$\angle ABC = 90˚$ (Right angled triangle)
$ \angle BAC = 90 - 44 = 46˚$ (From the horizontal to vertical line = 90˚)
$\angle ACB = 180 - (46 + 90) = 44˚$ (Angles in a triangle add to 180˚)
2. Find all the angles in $\triangle ACD$
$\angle CAD = 44 - 32 = 12˚$ (Values given to find the angle)
$\angle ACD = 180 - 44 = 136˚$(Angles on a line add to 180˚)
$\angle CDA = 180 - (136 + 12) = 32˚$ (Angles in a triangle add to 180˚)
3. Find the distance of $AC$ using the sine law + calculator
$\frac{2500}{sin44} = \frac{AC}{sin90}$
$\frac{2500sin90}{sin44} = AC$
$\frac{2500}{sin44} = AC$
$\frac{2500}{0.69466...} = AC$
$3598.9$ meters = $AC$
4. Find the distance of $CD$ using Step #3, sine law and your calculator
$\frac{3598.9}{sin32} = \frac{CD}{sin12}$
$\frac{3598.9sin12}{sin32} = CD$
$\frac{748.2516...}{sin32} = CD$
$\frac{748.2516...}{0.529919...} = CD$
$1412.0$ meters = $CD$
Therefore the distance between the two boats is $\approx 1412.0$ meters.
