Answer:
Hello,
Step-by-step explanation:
In the triangle ABG, using law of sin
[tex]\dfrac{sin(121^o)}{7} =\dfrac{sin(41^o)}{AG} =\dfrac{sin(18^o)}{BG} \\\\AG=\dfrac{sin(41^o)*7}{sin(121^o)} =5.358(km)\\\\BG=\dfrac{sin(18^o)*7}{sin(121^o)} =2.526(km)\\\\CG=5+BG=7.526(km)\\\\GD=CG*sin(31^o)=3.875(km)\\\\AD=AG+GD=9.233(km)\\\\CD=CG*cos(31^o)=6.451(km)\\\\AC=\sqrt{AD^2+CD^2} =11.263(km)\\[/tex]
