Answer
The distance between the ends of the wire on the ceiling is 15.8 feet
Work Step by Step
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
