## Trigonometry (11th Edition) Clone

Let $a = 12.2~ft$ Let $b = 15.0~ft$ Let angle $C = 70.3^{\circ}$ We can use the law of cosines to find $c$, the distance between the ends of the wire on the ceiling: $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(12.2~ft)^2+(15.0~ft)^2-(2)(12.2~ft)(15.0~ft)~cos~70.3^{\circ}}$ $c = \sqrt{250.46~ft^2}$ $c = 15.8~ft$ The distance between the ends of the wire on the ceiling is 15.8 feet