Chapter 7 - Applications of Trigonometric Functions - Section 7.3 The Law of Cosines - 7.3 Assess Your Understanding - Page 559: 25

$B= 90^\circ$ $A \approx67.4^\circ$ $C\approx 22.6^\circ$

1. Use the Law of Cosines, we have $B=cos^{-1}(\frac{(12)^2+(5)^2-(13)^2}{2(12)(5)})=90^\circ$ 2. Use the Law of Sines, $\frac{sinA}{12}=\frac{sin90^\circ}{13}$, thus $A=sin^{-1}(\frac{12sin90^\circ}{13})\approx67.4^\circ$ 3. We have $C\approx(180-90-67.4)=22.6^\circ$