Given a triangle with C=108°, a=10, b=6.5, find side c

Given a triangle with C=108°, a=10, b=6.5, find side c

$$c^2=a^2+b^2-2ab*\cos(108\ensurement{^{\circ}}) \\ c^2 = 10^2+6.5^2-2*10*6.5*\cos(108\ensurement{^{\circ}}) \\ c^2 = 100 +42.25 - 130 * (-0.30901699437) \\ c^2 = 142.25 + 40.1722092687\\ c^2 = 182.422209269 \quad | \qaud \sqrt{} \\ c= 13.5$$

Using the Law of Cosines, we have

C^2 =

  $${{\mathtt{10}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6.5}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{6.5}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{108}}^\circ\right)} = {\mathtt{182.422\: \!209\: \!268\: \!75}}$$

C =

  $${\sqrt{{\mathtt{182.422\: \!209\: \!268\: \!75}}}} = {\mathtt{13.506\: \!376\: \!615\: \!093\: \!701\: \!7}}$$

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