Answer
$A=115.5^{\circ},B=42.5^{\circ},C=22^{\circ}$
Work Step by Step
$a=832ft,b=623ft,c=345ft$
For angle A, as per Laws of cosines,
$a^{2}=b^{2}+c^{2}-2bc(cosA)$
$cosA=\frac{623^{2}+345^{2}-832^{2}}{2\times623\times345} = -0.431$
So, $A=115.5^{\circ}$
$b^{2}=c^{2}+a^{2}-2ca(cosB)$
$cosB=\frac{345^{2}+832^{2}-623^{2}}{2\times345\times832}=0.737$
$B=42.5^{\circ}$
Hence, $C=180^{\circ}-(A+B)=180^{\circ}-(115.5^{\circ}+42.5^{\circ})=180^{\circ}-158^{\circ}=22^{\circ}$
So, $A=115.5^{\circ},B=42.5^{\circ},C=22^{\circ}$
