## Algebra and Trigonometry 10th Edition

A = $91.80^{o}$ B = $44.10^{o}$ C = $44.10^{o}$
Note: The Standard Form of the Law of Cosines is: $a^{2} = b^{2} + c^{2} - 2bc(cosA)$ Note: The Alternative Form of the Law of Cosines is: $cosA = \frac{b^{2} + c^{2} - a^{2}}{2bc}$ To solve the triangle, we need to find A, B, and C. We can use the Alternative Form of the Law of Cosines for each of the angles. Finding A: $cosA = \frac{52.5^{2} + 52.5^{2} - 75.4^{2}}{2(52.5)(52.5)}$ A = $91.80^{o}$ Finding B: $cosB = \frac{75.4^{2} + 52.5^{2} - 52.5^{2}}{2(75.4)(52.5)}$ B = $44.10^{o}$ Finding C: $cosC = \frac{75.4^{2} + 52.5^{2} - 52.5^{2}}{2(75.4)(52.5)}$ C = $44.10^{o}$ In Total: A = $91.80^{o}$ B = $44.10^{o}$ C = $44.10^{o}$