Trigonometry (11th Edition) Clone

There are two possible triangles with the given parts. One triangle has the following values: $A = 53.23^{\circ}, B = 39.68^{\circ},$ and $C = 87.09^{\circ}$ $a = 29.81~m, b = 23.76~m,$ and $c = 37.16~m$ Another triangle has the following values: $A = 126.77^{\circ}, B = 39.68^{\circ},$ and $C = 13.55^{\circ}$ $a = 29.81~m, b = 23.76~m,$ and $c = 8.72~m$
We can use the law of sines to find the angle $A$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~B}{b}$ $sin~A = \frac{(29.81~m)~sin~(39.68^{\circ})}{23.76~m}$ $sin~A = 0.801$ $A = arcsin(0.801)$ $A = 53.23^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-53.23^{\circ}-39.68^{\circ}$ $C = 87.09^{\circ}$ We can find the length of side $c$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $c = \frac{b~sin~C}{sin~B}$ $c = \frac{(23.76~m)~sin~(87.09^{\circ})}{sin~39.68^{\circ}}$ $c = 37.16~m$ Note that we can also find another angle for A. $A = 180-53.23^{\circ} = 126.77^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-126.77^{\circ}-39.68^{\circ}$ $C = 13.55^{\circ}$ We can find the length of side $c$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $c = \frac{b~sin~C}{sin~B}$ $c = \frac{(23.76~m)~sin~(13.55^{\circ})}{sin~39.68^{\circ}}$ $c = 8.72~m$