Answer
$A=144.9^{\circ},C=18.8^{\circ},b=37.2m$.
Work Step by Step
$B=16.3^{\circ}, a=76.3m,c=42.8m$
From Laws of cosines,
$b^{2}=c^{2}+a^{2}-2ca(cosB)$
or, $b^{2}=(42.8)^{2}+(76.3)^{2}-2(42.8)(76.3)(cos16.3^{\circ})=1384.8$
or, $b=\sqrt (1384.8)=37.2m$
Again using Laws of cosines,
$a^{2}=b^{2}+c^{2}-2bc(cosA)$
$(76.3)^{2}=(37.2)^{2}+(42.8)^{2}-2(37.2)(42.8)(cosA)$
$cosA=-0.818$
$A=cos^{-1}(-0.818)=144.9^{\circ}$
Thus, $C=180^{\circ}-(A+B)=180^{\circ}-(144.9^{\circ}+16.3^{\circ})=18.8^{\circ}$
So, $A=144.9^{\circ},C=18.8^{\circ},b=37.2m$.
