1. **State the problem:** We need to derive the formula for the area of a triangle using two sides and the sine of the included angle, then apply it to find the area of a triangle with sides 7 cm, 8 cm, and included angle 60°. 2. **Derivation of the formula:** Consider a triangle with sides $a$ and $b$ enclosing an angle $\theta$. 3. Drop a perpendicular from the vertex opposite the angle $\theta$ to the base side, creating a height $h$. 4. The height $h$ can be expressed as $h = b \sin \theta$ because it is the opposite side of the right triangle formed. 5. The area $A$ of the triangle is given by: $$ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} a h $$ 6. Substitute $h = b \sin \theta$ into the area formula: $$ A = \frac{1}{2} a (b \sin \theta) = \frac{1}{2} ab \sin \theta $$ 7. **Apply the formula:** Given $a=7$ cm, $b=8$ cm, and $\theta=60^\circ$. 8. Calculate $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$. 9. Substitute values into the formula: $$ A = \frac{1}{2} \times 7 \times 8 \times 0.866 = 28 \times 0.866 = 24.248 $$ 10. **Final answer:** The area of the triangle is approximately $24.25$ square centimeters.