## Trigonometry (11th Edition) Clone

Let $c$ be the length of the wire on the upper side of the hill. Let $a = 15~ft$ Let $b = 30~ft$ Let $C$ be the angle between these two sides. Then $C = 90^{\circ}-20^{\circ}= 70^{\circ}$ We can use the law of cosines to find $c$: $c^2 = a^2+b^2-2ab~cos~C$ $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(15~ft)^2+(30~ft)^2-2(15~ft)(30~ft)~cos~70^{\circ}}$ $c = 817.18~ft^2$ $c = 28.6~ft$ The length of the wire on the upper side of the hill is 28.6 feet Let $d$ be the length of the wire on the lower side of the hill. Let $a = 15~ft$ Let $b = 30~ft$ Let $D$ be the angle between these two sides. Then $D = 90^{\circ}+20^{\circ}= 110^{\circ}$ We can use the law of cosines to find $d$: $d^2 = a^2+b^2-2ab~cos~D$ $d = \sqrt{a^2+b^2-2ab~cos~D}$ $d = \sqrt{(15~ft)^2+(30~ft)^2-2(15~ft)(30~ft)~cos~110^{\circ}}$ $d = 1432.82~ft^2$ $d = 37.9~ft$ The length of the wire on the lower side of the hill is 37.9 feet