As title suggests, we have a triangle $\triangle ABC$ with angles in an arithmetic progression, and sides $b$ and $c$ in a ratio of $\sqrt{3}:\sqrt{2}$. This is a problem I saw from a mathematics textbook in India. While I don't exactly remember the name I do know it was for grade 11-12.
I'm posting this here to see what the easiest possible way could be to solve it. I'm going to share my own approach here, please share your own as well as pointing out any flaws in mine if you spot any:
Here's my method:
So, since we know that $\angle A, \angle B$ and $\angle C$ are in arithmetic progression, we can say that the arithmetic mean between $A$ and $C$ is $B$:
$$B=\frac{A+C}{2}$$
$$2B=A+C$$
$$3B=A+B+C$$
$$3B=180^\circ$$
$$\angle B=60^\circ$$
Now we know that:
$$\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$$
$$\frac{c}{b}=\frac{\sin{C}}{\sin{B}}$$
$$\sin{C}=\sin{B}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$$
$$\sin{C}=\bigg(\frac{\sqrt{3}}{2}\bigg)\bigg(\frac{\sqrt{2}}{\sqrt{3}}\bigg)$$
$$\sin{C}=\frac{1}{\sqrt{2}}$$
Therefore, $\angle C=45^\circ$ and thus, $\angle A=180^\circ-45^\circ-60^\circ=75^\circ.$
Best Answer
Since angles are in an arithmetic progression, $(\angle B - \alpha) + \angle B + (\angle B +\alpha) =180^\circ$ then $\angle B= 60^\circ$
Let's say $b=\sqrt3, c=\sqrt2$.
Draw $AH \perp BC$ then $\angle BAH=30^\circ$and $AH=\frac{\sqrt6}{2}$.
Then in $\triangle AHC$, $HC=\frac{\sqrt6}{2}$ (Pytagorean theorem) and $AH=HC=\frac{\sqrt6}{2}$.
$\angle C = \angle CAH = 45^\circ$.
$\angle A = 30^\circ+ 45^\circ=75^\circ$.