Consider the following for the two (02) items that follow: Let ABC...

Let $AB = c$, $BC = a$, and $AC = b$.

Given that the triangle is right-angled at $B$, we have $b^2 = a^2 + c^2$.

We are also given $AB + AC = 3 \Rightarrow c + b = 3 \Rightarrow b = 3 - c$.

Substituting $b$ in the Pythagoras theorem:

$a^2 = (3 - c)^2 - c^2 = 9 - 6c + c^2 - c^2 = 9 - 6c$

The area of the triangle is $\Delta = \dfrac{1}{2}ac = \dfrac{1}{2}c\sqrt{9 - 6c}$.

To maximize the area, we can maximize its square:

$S = \Delta^2 = \dfrac{1}{4}c^2(9 - 6c) = \dfrac{9}{4}c^2 - \dfrac{3}{2}c^3$

Differentiating $S$ with respect to $c$ and equating to zero for maximum area:

$\dfrac{dS}{dc} = \dfrac{9}{2}c - \dfrac{9}{2}c^2 = 0$

$c(1 - c) = 0 \Rightarrow c = 1$ (since $c > 0$)

For $c = 1$, $b = 3 - 1 = 2$.

In the right-angled $\triangle ABC$, we have:

$\cos A = \dfrac{AB}{AC} = \dfrac{c}{b} = \dfrac{1}{2}$

Therefore, $\angle A = \dfrac{\pi}{3}$.

Answer: $\dfrac{\pi}{3}$