Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 303: 40

$x = 10.4~in$

Let the length of side $a = 12.0~in$ and let angle $A = 70.4^{\circ}$ Let angle $B = 54.8^{\circ}$. We can find the length of the side opposite angle $B$, which is the length of $x$: $\frac{x}{sin~B} = \frac{a}{sin~A}$ $x = \frac{a~sin~B}{sin~A}$ $x = \frac{(12.0~in)~sin~(54.8^{\circ})}{sin~(70.4^{\circ})}$ $x = 10.4~in$