## Trigonometry 7th Edition

$A=115.5^{\circ},B=42.5^{\circ},C=22^{\circ}$
$a=832ft,b=623ft,c=345ft$ 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, $A=115.5^{\circ},B=42.5^{\circ},C=22^{\circ}$