TITLE: How to find the angle between two straight line equations?
QUESTION [6 upvotes]: The Problem
There are two straight lines with equations as follow

y=-2x+10 and y=-3x+6

their point of intersection is (2,6) and i am asked to find the angle between them ?
A detailed and easy explanation would be appriciated 
Source and background
CIE/Cambridge International A and AS Level/Mathematics (9709)/Nov 2005 P1 Q9
(its part 3 i solved it a little before asking below is the whole question if needed)

REPLY [2 votes]: $$d_1:y=-2x+10\Rightarrow k_1=\tan\alpha_1=-2$$ and $$d_2:y=-3x+6\Rightarrow k_2=\tan\alpha_2=-3$$
then
$$\angle(d_1,d_2)=\alpha_1-\alpha_2=\arctan(\tan(\alpha_1-\alpha_2))=\arctan \frac{\tan\alpha_1-\tan\alpha_2}{1+\tan\alpha_1\tan\alpha_2}=$$
$$=\arctan\frac{k_1-k_2}{1+k_1k_2}=\arctan\frac{-2-(-3)}{1+(-2)(-3)}=\arctan\frac{1}{7}=0,14189..=8^{\circ} 07'48,37''$$
because
$$\arctan(\tan x)=x$$ and
$$\tan(\alpha_1-\alpha_2)=\frac{\tan\alpha_1-\tan\alpha_2}{1+\tan\alpha_1\tan\alpha_2}$$