Question: Solve for c.
3
120°
c
10
C = [ ? ]
Law of Cosines: c² = a² + b² - 2ab \cdot \cos(C)
Round your answer to the nearest hundredth.
1. **State the problem:** We need to find the length of side $c$ in a triangle where two sides are $3$ and $10$, and the included angle between them is $120^\circ$.
2. **Formula:** Use the Law of Cosines:
$$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$
where $a=3$, $b=10$, and $C=120^\circ$.
3. **Substitute values:**
$$c^2 = 3^2 + 10^2 - 2 \times 3 \times 10 \times \cos(120^\circ)$$
4. **Calculate squares:**
$$c^2 = 9 + 100 - 60 \cdot \cos(120^\circ)$$
5. **Evaluate cosine:** Recall that $\cos(120^\circ) = -\frac{1}{2}$.
$$c^2 = 109 - 60 \times \left(-\frac{1}{2}\right)$$
6. **Simplify multiplication:**
$$c^2 = 109 + 30$$
7. **Add:**
$$c^2 = 139$$
8. **Take square root:**
$$c = \sqrt{139}$$
9. **Calculate approximate value:**
$$c \approx 11.79$$
**Final answer:** $c \approx 11.79$ (rounded to the nearest hundredth).