If A = <5 2 8 > B = <6 5 3 > and C=A-B what is the angle between A and C?

I got ##59.45^@## (using Cosine formula …see A.S.Adikesavan’s answer using Sine formula)
My understanding is that:
##color(blue)——————————————————-##
##color(blue)(| )Given vectors: ##
##color(blue)(| )color(white)(XXX)color(blue)( vecu=< u_1u_2u_3 > andcolor(white)(XXXXX)color(blue)(|##
##color(blue)(| )color(white)(XXX)color(blue)(vecv = < v_1v_2v_3>color(white)(XXXXXxXXX)color(blue)(|)##
##color(blue)(| )##For an angle ##color(blue)(theta)## between ##color(blue)(vecu)## and ##color(blue)(vecv)####color(white)(XXXXxX)color(blue)(|)##
##color(blue)(| ) color(white)(XXX)color(blue)(sin(theta)=(vecu * vecv)/(abs(vecu)*abs(vecv))color(white)(XXXXXXXxXX)color(blue)(|)##
##color(blue)(| )##where##color(white)(XXXXXXXXXXXX)##
##color(blue)(| )color(white)(XXX)color(blue)(vecu * vecv = u_1 * v_1+u_2 * v_2 + u_3 * v_3)color(white)(X)color(blue)(|)##
##color(blue)(| )color(white)(XXX)color(blue)(abs(vecu)=sqrt(u_1^2+u_2^2+u_3^2))color(white)(XXXXXXxX)color(blue)(|)##
##color(blue)(| )color(white)(XXX)color(blue)(abs(vecv)=sqrt(v_1^2+v_2^2+v_3^2))color(white)(XXXXXXXX)color(blue)(|)##
##color(blue)——————————————————-##
In this case we are given
##color(white)(XXX)color(red)(vecA= < 528 >) color(red)(vecB= < 653>) and color(red)(vecC=vecA-vecB)##
##color(white)(XXX)rarr color(red)(vecC=< -1-35 >)##
##color(red)(vecA) * color(red)(vecC)=(5)(-1)+(2)(-3)+(8)(5) color(red)(= 29)##
##color(red)(abs(vecA))=sqrt(5^2+2^2+8^2)color(red)(=sqrt(93))##
##color(red)(abs(vecC))=sqrt(1^1+3^2+5^2)color(red)(=sqrt(35))##
Therefore
##color(white)(XXX)color(green)(cos(theta))=color(red)(29)/(color(red)(sqrt(93) * sqrt(35)))##
Using a calculator:
##color(white)(XXX)color(green)(cos(theta))=0.508303##
and
##color(white)(XXX)color(green)(theta) = arccos(0.508303) = color(green)(59.45^@)##