Answer:
b=12 , a=6[tex]6 \sqrt{3}[/tex] , c=6[tex]\sqrt{2}[/tex]
Step-by-step explanation:
based on the graph you are showing, you can use "SOH CAH TOA"
for right triangles, then you use "CAH" for get b:
[tex]Cos(60)=\frac{6}{b}\\b*Cos(60)=6\\b*\frac{1}{2}=6\\ b=6*2\\b=12\\\\[/tex]
you do the same for a, but in this case you use sin, not cos:
[tex]sin(60)=\frac{a}{b} \\b*sin(60)=a\\\\12*\sqrt{3}/2=a\\ 6\sqrt{3}=a\\[/tex]
and with your b value, you can get c, but now you use Cos with the 45 angle:
[tex]b*cos(45)=c\\12*\sqrt{2}/2=c\\ 6\sqrt{2}=c[/tex]
remember SOH CAH TOA means, Sin(x)=opposite/Hypotenuse, Cos(x)=adjacent/hypotenuse, and tan(x)=Opposite/adjacent.
