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.!