Trigonometry (11th Edition) Clone

The angle opposite the side of length 45 feet is $26.4^{\circ}$ The angle opposite the side of length 60 feet is $36.3^{\circ}$
Let $a = 45~ft$, let $b = 60~ft$, and let $c = 90~ft$. We can use the law of cosines to find $A$, which is the angle which subtends the side $a$: $a^2 = b^2+c^2-2bc~cos~A$ $2bc~cos~A = b^2+c^2-a^2$ $cos~A = \frac{b^2+c^2-a^2}{2bc}$ $A = arccos(\frac{b^2+c^2-a^2}{2bc})$ $A = arccos(\frac{60^2+90^2-45^2}{(2)(60)(90)})$ $A = arccos(0.89583)$ $A = 26.4^{\circ}$ The angle opposite the side of length 45 feet is $26.4^{\circ}$ We can use the law of cosines to find $B$, which is the angle which subtends the side $b$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{45^2+90^2-60^2}{(2)(45)(90)})$ $B = arccos(0.8056)$ $B = 36.3^{\circ}$ The angle opposite the side of length 60 feet is $36.3^{\circ}$