16*16 = 11*11 +12*12 - 2(11)(12)Cos-1(

Looks like you are trying to use the Cosine rule to find an angle (I'll call it A) of a triangle given the lengths of the three sides.  In general, if the sides are a, b and c (with 'a' opposite angle A) then

a² = b² + c² -2*b*c*cos(A) or, rearranging this to get cos(A)

cos(A) = ( b² + c² - a²)/(2*b*c)

Here you have a = 16, b = 11, c = 12 so

$${\mathtt{cosA}} = {\frac{\left({{\mathtt{11}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{12}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{16}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)}} \Rightarrow {\mathtt{cosA}} = {\mathtt{0.034\: \!090\: \!909\: \!090\: \!909\: \!1}}$$

and 

$${\mathtt{A}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{\left({{\mathtt{11}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{12}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{16}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)}}\right)} \Rightarrow {\mathtt{A}} = {\mathtt{88.046\: \!356\: \!247\: \!078^{\circ}}}$$