Ans: 5.36 cm² (5.360 to 5.361)
Given the triangle ABC with sides \(a = 4.9 \, \text{cm}\), \(b = 5.6 \, \text{cm}\), and included angle \(C = 23^\circ\), the area is calculated using the formula:
\[ \text{Area} = \frac{1}{2}ab \sin C \]
Substitute the known values:
\[ \text{Area} = \frac{1}{2} \times 4.9 \times 5.6 \times \sin(23^\circ) \]
First, compute \(\sin(23^\circ) \approx 0.3907\).
Now, multiply the values:
\[ \text{Area} = \frac{1}{2} \times 4.9 \times 5.6 \times 0.3907 \approx \frac{1}{2} \times 10.717 \approx 5.36 \, \text{cm}^2 \]
Thus, the area of triangle ABC is approximately 5.36 cm².
