1. Stating the problem: Given a triangle ABC with perimeter 15 cm, side AB = 7 cm, and angle BAC = 60 degrees, find the lengths of AC and BC and the area of the triangle. 2. Define variables: Let AC = x cm and BC = y cm. 3. Use the perimeter formula: AB + BC + AC = 15 Substitute known values: 7 + y + x = 15 Simplify: y + x = 8 So, $$y = 8 - x$$. 4. By the Law of Cosines for angle BAC (60°): $$ BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(60^\circ) $$ Substitute values and variables: $$ y^2 = 7^2 + x^2 - 2 \times 7 \times x \times \frac{1}{2} $$ Simplify: $$ y^2 = 49 + x^2 - 7x $$ 5. Substitute $$y = 8 - x$$ into the equation: $$ (8 - x)^2 = 49 + x^2 - 7x $$ Expand left side: $$ 64 - 16x + x^2 = 49 + x^2 - 7x $$ 6. Cancel $$x^2$$ from both sides: $$ 64 - 16x = 49 - 7x $$ Bring variables to one side: $$ -16x + 7x = 49 - 64 $$ Simplify: $$ -9x = -15 $$ Divide both sides by -9: $$ x = \frac{15}{9} = \frac{5}{3} \approx 1.67 \text{ cm} $$ 7. Find $$y$$: $$ y = 8 - x = 8 - \frac{5}{3} = \frac{24}{3} - \frac{5}{3} = \frac{19}{3} \approx 6.33 \text{ cm} $$ 8. Calculate the area of the triangle using formula: $$ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(60^\circ) $$ Substitute known values: $$ = \frac{1}{2} \times 7 \times \frac{5}{3} \times \frac{\sqrt{3}}{2} $$ Simplify: $$ = \frac{35}{6} \times \frac{\sqrt{3}}{2} = \frac{35 \sqrt{3}}{12} \approx 5.05 \text{ cm}^2 $$ Final answers: - AC = $$\frac{5}{3}$$ cm approximately 1.67 cm - BC = $$\frac{19}{3}$$ cm approximately 6.33 cm - Area = $$\frac{35 \sqrt{3}}{12}$$ cm² approximately 5.05 cm²