How do you solve the triangle given ∠B = 99°, ∠C = 47°, b = 23?

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Answer 1 · 2016-08-21T14:35:15

in #DeltaABC#, # m/_A=34^@, a=13.02, c=17.03#.

#m/_B+m/_C=99^@+47^@=146^@#.

#:. m/_A=180^@-146^@=34^@#.

Now, we know the Sine-Rule (in usual notation) from Trigo. :

In #DeltaABC, a/sinA=b/sinB=c/sinC#.

#:. a/(sin34^@)=23/(sin99^@)=c/(sin47^@)#.

#:. a=(23sin34^@)/(sin99^@)=(23)(0.5592)/0.9877=13.02#.

#rArr c=(23sin47^@)/(sin99^@)=(23)(0.7314)/0.9877=17.03#

Thus, in #DeltaABC#, # m/_A=34^@, a=13.02, c=17.03#.

Enjoy Maths.!