Find the angles of the triangle whose sides are x + y – 4 = 0, 2x + y – 6 = 0 and 5x + 3y – 15 = 0.

Given lines
$\begin{array}{l}\mathrm{x}+\mathrm{y}-4=0 .....\left(1\right)\\ 2\mathrm{x}+\mathrm{y}-6=0 ....\left(2\right)\\ 5\mathrm{x}+3\mathrm{y}-15=0 ...\left(3\right)\end{array}$
The angle b/w (1) & (2) 
$\begin{array}{l}\mathrm{the} \mathrm{angle} \mathrm{between} {\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\text{ and }{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0\text{ is }\\ \\ \mathrm{cos\theta }=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2}}\\ \cos (\mathrm{\pi }-\mathrm{\theta })=\frac{2+1}{\sqrt{2}\sqrt{4+1}}\\ \Rightarrow \mathrm{\pi }-\mathrm{\theta }={\cos }^{-1}\left(\frac{3}{\sqrt{2}\cdot \sqrt{5}}\right)\Rightarrow \mathrm{\theta }=\mathrm{\pi }-{\cos }^{-1}\left(\frac{3}{\sqrt{10}}\right)\end{array}$
The angle b/w (2) & (3) is
$\begin{array}{l}\cos \mathrm{\theta }=\frac{2.5+1.3}{\sqrt{4+1}\sqrt{25+9}}=\frac{10+3}{\sqrt{5}\sqrt{34}}=\frac{13}{\sqrt{170}}\\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{13}{\sqrt{170}}\right)\end{array}$
The angle b/w (3) & (1) is
$\begin{array}{l}\cos \mathrm{\theta }=\frac{5.1+3.1}{\sqrt{2}\sqrt{34}}=\frac{8}{2\sqrt{17}}=\frac{4}{\sqrt{17}}\\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{4}{\sqrt{17}}\right)\end{array}$