The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be $$\alpha$$. After moving a distance 2 metres from P towards the foot of the tower, the angle of elevation changes to $$\beta$$. Then the height (in metres) of the tower is:

Let the height of the tower be $$ h $$ meters. Let the distance from the new point (after moving 2 meters towards the tower) to the foot of the tower be $$ x $$ meters. Therefore, the distance from the original point P to the foot of the tower is $$ x + 2 $$ meters.

In the right triangle formed by point P, the foot of the tower, and the top of the tower, the angle of elevation is $$ \alpha $$. The opposite side to angle $$ \alpha $$ is the height $$ h $$, and the adjacent side is $$ x + 2 $$. Using the tangent function:

$$ \tan \alpha = \frac{h}{x + 2} $$

In the right triangle formed by the new point (after moving 2 meters), the foot of the tower, and the top of the tower, the angle of elevation is $$ \beta $$. The opposite side to angle $$ \beta $$ is $$ h $$, and the adjacent side is $$ x $$. Using the tangent function:

$$ \tan \beta = \frac{h}{x} $$

From the second equation, express $$ h $$ in terms of $$ x $$:

$$ h = x \tan \beta $$

Substitute this expression for $$ h $$ into the first equation:

$$ \tan \alpha = \frac{x \tan \beta}{x + 2} $$

Solve for $$ x $$. Multiply both sides by $$ x + 2 $$:

$$ \tan \alpha \cdot (x + 2) = x \tan \beta $$

Expand the left side:

$$ x \tan \alpha + 2 \tan \alpha = x \tan \beta $$

Bring terms containing $$ x $$ to one side:

$$ x \tan \alpha - x \tan \beta = -2 \tan \alpha $$

Multiply both sides by -1 to simplify:

$$ x \tan \beta - x \tan \alpha = 2 \tan \alpha $$

Factor out $$ x $$:

$$ x (\tan \beta - \tan \alpha) = 2 \tan \alpha $$

Solve for $$ x $$:

$$ x = \frac{2 \tan \alpha}{\tan \beta - \tan \alpha} $$

Recall that $$ h = x \tan \beta $$. Substitute the expression for $$ x $$:

$$ h = \left( \frac{2 \tan \alpha}{\tan \beta - \tan \alpha} \right) \tan \beta $$

Simplify:

$$ h = \frac{2 \tan \alpha \tan \beta}{\tan \beta - \tan \alpha} $$

Express tangent in terms of sine and cosine:

$$ \tan \alpha = \frac{\sin \alpha}{\cos \alpha}, \quad \tan \beta = \frac{\sin \beta}{\cos \beta} $$

Substitute:

$$ h = \frac{2 \cdot \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta}}{\frac{\sin \beta}{\cos \beta} - \frac{\sin \alpha}{\cos \alpha}} $$

Simplify the denominator:

$$ \frac{\sin \beta}{\cos \beta} - \frac{\sin \alpha}{\cos \alpha} = \frac{\sin \beta \cos \alpha - \sin \alpha \cos \beta}{\cos \alpha \cos \beta} $$

Recognize the numerator as the sine difference formula:

$$ \sin \beta \cos \alpha - \sin \alpha \cos \beta = \sin(\beta - \alpha) $$

So the denominator becomes:

$$ \frac{\sin(\beta - \alpha)}{\cos \alpha \cos \beta} $$

Now substitute back into the expression for $$ h $$:

$$ h = \frac{2 \cdot \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\sin(\beta - \alpha)}{\cos \alpha \cos \beta}} $$

Dividing fractions is equivalent to multiplying by the reciprocal:

$$ h = \frac{2 \sin \alpha \sin \beta}{\cos \alpha \cos \beta} \cdot \frac{\cos \alpha \cos \beta}{\sin(\beta - \alpha)} $$

Cancel $$ \cos \alpha \cos \beta $$ in numerator and denominator:

$$ h = \frac{2 \sin \alpha \sin \beta}{\sin(\beta - \alpha)} $$

This matches Option A. The other options do not match this expression. Therefore, the height of the tower is $$ \frac{2 \sin \alpha \sin \beta}{\sin(\beta - \alpha)} $$ meters.

Hence, the correct answer is Option A.