You have the triangle ABC. Angle A = 40, side c = 7, and side b = 6. Using the Law of Cosines, find angles A and B.

Since you are given angle A, I suspect you mean angles B and C!

Let 'a' be the length of the side from B to C.  The law of cosines says:

a² = b² + c² - 2*b*c*cos(A)

Put the known information in here:

a² = 6² + 7² -2*6*7*cos(40°)

so 'a' is the square root of the right-hand side:

$${\mathtt{a}} = {\sqrt{{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{7}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{7}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{40}}^\circ\right)}}} = {\mathtt{a}} = {\mathtt{4.544\: \!476\: \!513\: \!087\: \!508\: \!5}}$$

Now we can use the cosine rule in a rearranged form to get cos(B):

cos(B) = (a² + c² - b²)/(2*a*c) or B = cos^-1((a² + c² - b²)/(2*a*c))

$${\mathtt{B}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{\left({{\mathtt{4.544\: \!476\: \!513}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{7}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{6}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{4.544\: \!476\: \!513}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)}}\right)} = {\mathtt{B}} = {\mathtt{58.066\: \!464\: \!967\: \!59^{\circ}}}$$

You could use the cosine rule again changing 'b' for 'c' to find angle C if you like.  However, it's probably easier to get it by knowing that A + B + C = 180°

$${\mathtt{C}} = {\mathtt{180}}{\mathtt{\,-\,}}{\mathtt{40}}{\mathtt{\,-\,}}{\mathtt{58.066\: \!465}} = {\mathtt{C}} = {\mathtt{81.933\: \!535}}$$ °

Also, having obtained the size of 'a' it might have been easier to use the sine rule to find B (sin(B)/b = sin(A)/a).