How do you solve a triangle given C=16 degrees c=32 b=92?

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Answer 1 · 2018-05-21T06:51:32

#A=111.6^@#
#B=52.4^@#
#a=107.9#

#[sinC]/c=[sinB]/b#

#[sin16]/32=[sinB]/92#

#sin B=92 xx [sin16]/32#

#sin B=0.792457398#

#B=sin^(-1) (0.792457398)#

B=52.41575584

#B=52.4^@ => A=111.6^@#

#32/[sin16]=a/[sin111.6]#

#a=32/[sin16] xx sin111.6#

#a=107.9420003#