Angle C Measure 6628Ee

1. **State the problem:** We need to find the measure of angle $C$ in a triangle with sides $a=55$, $b=50$, and $c=90$. 2. **Formula used:** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$ This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. 3. **Substitute known values:** $$90^2 = 55^2 + 50^2 - 2 \times 55 \times 50 \cdot \cos(C)$$ Calculate squares: $$8100 = 3025 + 2500 - 5500 \cdot \cos(C)$$ 4. **Simplify the right side:** $$8100 = 5525 - 5500 \cdot \cos(C)$$ 5. **Isolate the cosine term:** $$8100 - 5525 = -5500 \cdot \cos(C)$$ $$2575 = -5500 \cdot \cos(C)$$ 6. **Divide both sides by $-5500$:** $$\frac{2575}{\cancel{5500}} = \frac{-5500 \cdot \cos(C)}{\cancel{5500}}$$ $$\frac{2575}{5500} = -\cos(C)$$ 7. **Simplify fraction:** $$\frac{2575}{5500} = \frac{2575 \div 25}{5500 \div 25} = \frac{103}{220} \approx 0.46818$$ So, $$0.46818 = -\cos(C)$$ 8. **Solve for $\cos(C)$:** $$\cos(C) = -0.46818$$ 9. **Find angle $C$ using inverse cosine:** $$C = \cos^{-1}(-0.46818)$$ Calculate: $$C \approx 118.00^\circ$$ 10. **Final answer:** The measure of angle $C$ is approximately **118.00°** rounded to the nearest hundredth.