In triangle ABC, AB = 45cm, AC = 29.5cm, BC = 35.3cm and angle CAB = 51.6°. (a) Calculate angle ABC. (b) Calculate the area of triangle ABC.

The Sine Rule states that for a triangle with sides $a, b, c$ and opposite angles $A, B, C$:

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

In $\triangle ABC$, we have:
Side $a = BC = 35.3$ cm
Angle $A = \angle CAB = 51.6^\circ$
Side $b = AC = 29.5$ cm
Angle $B = \angle ABC$ (which we need to find)

Using the Sine Rule:

$$\frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)}$$ $$\frac{35.3}{\sin 51.6^\circ} = \frac{29.5}{\sin(\angle ABC)}$$

Cross-multiply to isolate $\sin(\angle ABC)$:

$$35.3 \times \sin(\angle ABC) = 29.5 \times \sin 51.6^\circ$$ $$\sin(\angle ABC) = \frac{29.5 \times \sin 51.6^\circ}{35.3}$$ $$\sin(\angle ABC) = \frac{29.5 \times 0.783636...}{35.3}$$ $$\sin(\angle ABC) = \frac{23.11726...}{35.3}$$ $$\sin(\angle ABC) = 0.65488...$$

$$\angle ABC = \sin^{-1}(0.65488...)$$ $$\angle ABC \approx 40.900^\circ$$

Rounding to one decimal place:

$$\angle ABC = \boxed{40.9^\circ}$$

The area of a triangle can be calculated using the formula:

$$\text{Area} = \frac{1}{2}ab \sin C$$

where $a$ and $b$ are the lengths of two sides, and $C$ is the included angle between those sides.
In $\triangle ABC$, we can use sides $AC$ and $AB$ and the included angle $\angle CAB$:
Side $AC = 29.5$ cm
Side $AB = 45$ cm
Included angle $\angle CAB = 51.6^\circ$

$$\text{Area} = \frac{1}{2} \times AC \times AB \times \sin(\angle CAB)$$ $$\text{Area} = \frac{1}{2} \times 29.5 \text{ cm} \times 45 \text{ cm} \times \sin(51.6^\circ)$$ $$\text{Area} = \frac{1}{2} \times 1327.5 \text{ cm}^2 \times 0.783636...$$ $$\text{Area} = 663.75 \text{ cm}^2 \times 0.783636...$$ $$\text{Area} \approx 519.063 \text{ cm}^2$$

Rounding to three significant figures:

$$\text{Area} = \boxed{519 \text{ cm}^2}$$