## Trigonometry (11th Edition) Clone

There are two possible triangles for these values. One triangle is as follows: $A = 51.20^{\circ}~$, $~B = 69.09^{\circ}~$, $~C=59.71^{\circ}$ $a = 7208~cm~$, $~b = 8640~cm~$, $c = 7986~cm$ The other triangle is as follows: $A = 51.20^{\circ}~$, $~B = 8.51^{\circ}~$, $~C=120.29^{\circ}$ $a = 7208~cm~$, $~b = 1369~cm~$, $c = 7986~cm$
We can use the law of sines to find the angle $C$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $sin~C = \frac{c~sin~A}{a}$ $sin~C = \frac{(7986~cm)~sin~(51.20^{\circ})}{7208~cm}$ $sin~C = 0.8635$ $C = arcsin(0.8635)$ $C = 59.71^{\circ}$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-51.20^{\circ}-59.71^{\circ}$ $B = 69.09^{\circ}$ We can find the length of side $b$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $b = \frac{a~sin~B}{sin~A}$ $b = \frac{(7208~cm)~sin~(69.09^{\circ})}{sin~(51.20^{\circ})}$ $b = 8640~cm$ Note that we can also find another angle for C. $C = 180-59.71^{\circ} = 120.29^{\circ}$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-51.20^{\circ}-120.29^{\circ}$ $B = 8.51^{\circ}$ We can find the length of side $b$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $b = \frac{a~sin~B}{sin~A}$ $b = \frac{(7208~cm)~sin~(8.51^{\circ})}{sin~(51.20^{\circ})}$ $b = 1369~cm$