## Trigonometry (11th Edition) Clone

There are two possible triangles with the given parts. One possible triangle has the following values: $A = 55^{\circ}20', B = 94^{\circ}50'$, and $C = 29^{\circ}50'$ $a = 8.61~m, b = 10.44~m$, and $c = 5.21~m$ Another possible triangle has the following values: $A = 124^{\circ}40', B = 25^{\circ}30'$, and $C = 29^{\circ}50'$ $a = 8.61~m, b = 4.51~m$, and $c = 5.21~m$
We can use the law of sines to find the angle $A$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~C}{c}$ $sin~A = \frac{(8.61~m)~sin~(29^{\circ}50')}{5.21~m}$ $A = arcsin(0.8221)$ $A = 55^{\circ}20'$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-55^{\circ}20'-29^{\circ}50'$ $B = 94^{\circ}50'$ We can find the length of side $b$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $b = \frac{c~sin~B}{sin~C}$ $b = \frac{(5.21~m)~sin~(94^{\circ}50')}{sin~29^{\circ}50'}$ $b = 10.44~m$ Note that we can also find another angle for A. $A = 180-55^{\circ}20' = 124^{\circ}40'$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-124^{\circ}40'-29^{\circ}50'$ $B = 25^{\circ}30'$ We can find the length of side $b$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $b = \frac{c~sin~B}{sin~C}$ $b = \frac{(5.21~m)~sin~(25^{\circ}30')}{sin~29^{\circ}50'}$ $b = 4.51~m$