find angle between the line $y = x + 3 \text { and } y = 2 x + 3$ hello i tried to solve but i am not sure if its accurate answer 
that how i did it by using this formula
$\tan \theta = \frac { m _ { 2 } - m _ { 1 } } { 1 + m _ { 1 } m _ { 2 } }$
1st line $y=x+3$
assumed $1$ as $m _ { 1 }$
2nd line $y = 2 x + 3y$
$m _ { 2 }=2$
applied values to formula
$= \frac { 2 - 1 } { 1 + ( 1 ) ( 2 ) } = \frac { 1 } { 1 + 2 } = \frac { 1 } { 3 }$
$\tan^{-1} (\frac { 1 } { 3 }) = 18.43494°$
i am kind of curious if this is right or i did something wrong.

Yes it is correct, as an alternative we have that


*

*$y=x+3$ has direction vector $u=(1,1)$

*$y=2x+3$ has direction vector $v=(1,2)$


therefore by dot product
$$\cos \theta=\frac{u\cdot v}{|u||v|}=\frac{3}{\sqrt{10}}\implies \theta =\arccos \frac{3}{\sqrt{10}}=18.43°$$
Note also that since $\sin \theta = \sqrt{1-\cos^2 \theta}$ we have
$$\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\sqrt{1-\cos^2 \theta}}{\cos \theta}=\frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}}=\frac13$$
which agrees with your result.
