## Trigonometry (11th Edition) Clone

There are two possible triangles with the given parts. One triangle has the following values: $A = 38^{\circ}40', B = 49^{\circ}20',$ and $C = 92^{\circ}$ $a = 9.72~m, b = 11.8~m$, and $c = 15.5~m$ Another triangle has the following values: $A = 38^{\circ}40', B = 130^{\circ}40',$ and $C = 10^{\circ}40'$ $a = 9.72~m, b = 11.8~m$, and $c = 2.88~m$
We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(11.8~m)~sin~(38^{\circ}40')}{9.72~m}$ $B = arcsin(0.7585)$ $B = 49^{\circ}20'$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-38^{\circ}40'-49^{\circ}20'$ $C = 92^{\circ}$ We can find the length of side $c$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(9.72~m)~sin~(92^{\circ})}{sin~38^{\circ}40'}$ $c = 15.5~m$ Note that we can also find another angle for B. $B = 180-49^{\circ}20' = 130^{\circ}40'$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-38^{\circ}40'-130^{\circ}40'$ $C = 10^{\circ}40'$ We can find the length of side $c$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(9.72~m)~sin~(10^{\circ}40')}{sin~38^{\circ}40'}$ $c = 2.88~m$