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Sagot :
[tex]\frac{|AC|}{sin\angle ABC}=\frac{|BC|}{sin\angle BAC}\\\\|AC|=19cm;\ |BC|=17cm;\ |\angle BAC|=60^o;\ |\angle ABC|=x\\\\\frac{19}{sinx}=\frac{17}{sin60^o}\\\\cross\ multiply\\\\17sinx=19sin60^O\\\\17sinx=19\cdot\frac{\sqrt3}{3}\ \ \ \ /:17\\\\sinx=\frac{19\sqrt3}{51}\\\\sinx\approx0.6453\Rightarrow x\approx40^o11'\\\\Answer:|\angle ABC|\approx40^o11'[/tex]
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