In triangle ABC,$${\mathtt{AB}} = {\sqrt{{\mathtt{2}}}}$$,point D is on side BC, BD=2*DC,cosDAC=$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}$$,cosC=$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}$$,AC+BC=?
$$\small{
\begin{array}{l}
\angle DAC = \arccos(\frac{3\cdot\sqrt{10}}{10}) = 18.4349488229\ensurement{^{\circ}}\\\\
\angle C = \arccos(\frac{2\cdot\sqrt{5}}{5}) = 26.5650511771\ensurement{^{\circ}}\\\\
\angle DAC + \angle C = 45 \ensurement{^{\circ}} \\\\
\angle ADC = 180 \ensurement{^{\circ}} -45 \ensurement{^{\circ}}
= 135\ensurement{^{\circ}}\\\\
\angle BDA = 180 \ensurement{^{\circ}} -135 \ensurement{^{\circ}}
= 45\ensurement{^{\circ}}\\\\
\end{array}
}$$
Now we define:
DC = x, BD = 2x,
H = foot (of a perpendicular) from A on line BC between B and D. ($$\small{\angle BDA = 45\ensurement{^{\circ}}}$$ !!!)
h = AH
u = BH, v = HD, DB = u+v = 2x
$$\small{
\begin{array}{l}
H = \text{foot (of a perpendicular) from }~ A \text{ to line $\overline{BC}$ } \\
h = \overline{AH}\qquad
u = \overline{BH}\qquad
v = \overline{HD}\qquad
x = \overline{DC}\qquad
\overline{BC} = 3x
\end{array}
}\\\\
\small{
\begin{array}{rcl}
\tan{(45\ensurement{^{\circ}} )}=1 &=& \frac{h}{v} \\
h &=& v\\\\
\tan{(C)} = 0.5 &=& \frac{h}{v+x} \\
0.5 &=& \frac{v}{v+x} \\
v+x &=& 2v\\
v &=& x\\\\
\overline{BD} = u+v &=& 2x \\
u&=& 2x-v \\
u &=& 2x -x \\
u &=& x\\\\
u^2+h^2 &=& \sqrt{2}^2 \quad | \quad u= 2x-v\\
(2x-v)^2 + v^2 &=& 2 \quad | \quad v=x\\
(2x-x)^2 + x^2 &=& 2 \\
x^2+x^2 &=& 2\\
2x^2 &=& 2 \\
x^2 &=& 1\\
x &=& 1\\\\
\overline{BC} = 3x = 3\cdot 1 = 3\\\\
\frac {\overline{AC}}
{ \sin{(135\ensurement{^{\circ}})} }&=&\frac{x}{\sin{(DAC)}}\\\\
\frac {\overline{AC}} {\sin{(135\ensurement{^{\circ}} )}} &=&\frac{1}{ \sin{(DAC)} }\\\\
\overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{(DAC)} } \\\\
\end{array}
}\\\\$$
$$\small{
\begin{array}{rcl}
\overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}}
{ \sin{ ( 18.4349488229 \ensurement{^{\circ}} )} } \\\\
\overline{AC} &=& 2.23606797750 \\\\
\overline{AC}+\overline{BC}& =& 2.23606797750 +3\\\\
\mathbf{ \overline{AC}+\overline{BC}} &\mathbf{ =}& \mathbf{5.23606797750}
&\\
\hline
\end{array}
}$$
