How do you use law of sines to solve the triangle given C = 37°, a = 19, c = 8?

1 Answer
Nov 16, 2015

#A=115.4^o#
#B=26.6^o#
#b=6.16#

Explanation:

#(sinA)/a=(sinB)/b=(sinC)/c#

#(sinA)/19=(sinC)/8#
Multiply both sides by 19 and #sinA=19(sin37)/8# which is #sinA=1.429#
The next step is to take the #sin^-1# but you can not do an inverse of sin > 1 . We know that #sin^-1(1)=90^o# #sin^-1(0.429)=25.4# #90+25.4=115.4^o# That's your angle A.

Angle B is #180-115.4-37=27.6^o#

#b/(sin27.6)=8/(sin37# so #b=sin27.6(8/(sin37))#
#b=6.16#