In a triangle ABC, side a = 6.7 cm, side c = 2.3 cm and angle B = 46.6 degrees. Find side b and angle A.

The Cosine Rule states $b^2 = a^2 + c^2 - 2ac \cos B$.
Given $a = 6.7 \text{ cm}$, $c = 2.3 \text{ cm}$, and $B = 46.6^\circ$.

Substitute the values into the formula:

$$b^2 = (6.7)^2 + (2.3)^2 - 2(6.7)(2.3) \cos(46.6^\circ)$$

Calculate the squares:

$$b^2 = 44.89 + 5.29 - 2(6.7)(2.3) \cos(46.6^\circ)$$

Sum the squared terms:

$$b^2 = 50.18 - 2(6.7)(2.3) \cos(46.6^\circ)$$

Multiply the terms in the cosine part:

$$b^2 = 50.18 - 30.82 \cos(46.6^\circ)$$

Calculate $\cos(46.6^\circ)$:

$$\cos(46.6^\circ) \approx 0.687004$$

Substitute this value:

$$b^2 = 50.18 - 30.82(0.687004)$$ $$b^2 = 50.18 - 21.179999999999996$$ $$b^2 = 29.000000000000004$$

Take the square root to find $b$:

$$b = \sqrt{29.000000000000004}$$ $$b \approx 5.38516 \text{ cm}$$

Rounding to three significant figures:

$$b \approx 5.39 \text{ cm}$$

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B}$.
We have $a = 6.7 \text{ cm}$, $b \approx 5.38516 \text{ cm}$, and $B = 46.6^\circ$.

Rearrange the formula to solve for $\sin A$:

$$\sin A = \frac{a \sin B}{b}$$

Substitute the values:

$$\sin A = \frac{6.7 \sin(46.6^\circ)}{5.38516}$$

Calculate $\sin(46.6^\circ)$:

$$\sin(46.6^\circ) \approx 0.726591$$

Substitute this value:

$$\sin A = \frac{6.7 \times 0.726591}{5.38516}$$ $$\sin A = \frac{4.8681597}{5.38516}$$ $$\sin A \approx 0.90400$$

Now, find $A$ by taking the inverse sine:

$$A = \arcsin(0.90400)$$ $$A \approx 64.690^\circ$$

Rounding to one decimal place:

$$A \approx 64.7^\circ$$

The sum of angles in a triangle is $180^\circ$.

$$A + B + C = 180^\circ$$

We have $A \approx 64.690^\circ$ and $B = 46.6^\circ$.

Rearrange to solve for $C$:

$$C = 180^\circ - A - B$$

Substitute the values:

$$C = 180^\circ - 64.690^\circ - 46.6^\circ$$ $$C = 180^\circ - 111.290^\circ$$ $$C = 68.710^\circ$$

Rounding to one decimal place:

$$C \approx 68.7^\circ$$

The final answers are:

$$b \approx \boxed{5.39 \text{ cm}}$$ $$A \approx \boxed{64.7^\circ}$$ $$C \approx \boxed{68.7^\circ}$$