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1. a = 5,\ b = 7,\ C = 60^\circ

   Two sides and the included angle (SAS): find the third side with the law of cosines.

2. c^2 = a^2 + b^2 - 2ab\cos C = 39

   Law of cosines.

3. c = \sqrt{39} \approx 6.2450

4. P = a + b + c = \sqrt{39} + 12

   Perimeter.

5. s = \tfrac{P}{2} = \frac{\sqrt{39}}{2} + 6,\quad A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{35 \sqrt{3}}{4} \approx 15.155

   Heron's formula for the area.

6. \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3 \sqrt{39}}{26} \Rightarrow A \approx 43.898^\circ

   Law of cosines for angle A (opposite side a).

7. \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{\sqrt{39}}{26} \Rightarrow B \approx 76.102^\circ

   Law of cosines for angle B (opposite side b).

8. \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{1}{2} \Rightarrow C \approx 60.000^\circ

   Law of cosines for angle C (opposite side c).

9. A + B + C = 180.0^\circ

   The angles add to 180°: a acute, scalene triangle.