1. **State the problem:** We have triangle ABC with a right angle at C, angle A = 120°, side AC = 8, and side BC = 5. We need to find the area $S_{ABC}$. 2. **Recall the formula for the area of a triangle:** $$S = \frac{1}{2} \times \text{base} \times \text{height}$$ 3. **Identify the base and height:** Since there is a perpendicular from B to AC at M, AC can be the base, and BM is the height. 4. **Find length BM (height):** Use the Pythagorean theorem in right triangle BMC: $$BM = \sqrt{BC^2 - MC^2}$$ 5. **Find MC:** Use the Law of Cosines in triangle ABC to find side AB: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(120^\circ)$$ Calculate: $$AB^2 = 8^2 + 5^2 - 2 \times 8 \times 5 \times \cos(120^\circ) = 64 + 25 - 80 \times (-0.5) = 89 + 40 = 129$$ So, $$AB = \sqrt{129}$$ 6. **Find MC using projection:** Since M is the foot of the perpendicular from B to AC, MC is the projection of BC onto AC. Using the formula for projection: $$MC = BC \times \cos(30^\circ)$$ Because angle at A is 120°, angle between BC and AC is $180^\circ - 120^\circ = 60^\circ$, so angle between BC and perpendicular from B to AC is $30^\circ$. Calculate: $$MC = 5 \times \cos(30^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$$ 7. **Calculate BM:** $$BM = \sqrt{5^2 - \left(\frac{5\sqrt{3}}{2}\right)^2} = \sqrt{25 - \frac{75}{4}} = \sqrt{\frac{100}{4} - \frac{75}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$ 8. **Calculate area:** $$S = \frac{1}{2} \times AC \times BM = \frac{1}{2} \times 8 \times \frac{5}{2} = 10$$ **Final answer:** $$S_{ABC} = 10$$