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
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)
On
Try using this relationship. If a line has slope $m$, and the angle at which it hits the $x$-axis is $\theta$, then $$\tan(\theta) = m.$$
On
The slope of a line is equal to the tan of its angle of inclination (angle above the positive x axis. So you can calculate that angle for each line and then find the difference. That is my initial suggestion.
On
The slope of $y=-2x+10$ is $-2$, so it makes an angle of $\arctan (-2)$ with the $x$ axis. Similarly the other has a slope of $-3$ and makes an angle of $\arctan (-3)$ with the $x$ axis. The angle between them is $\arctan (-2)-\arctan(-3)\approx 0.1419$ radians.
Another approach is that a vector in the direction of the first line is $(1,-2)$ and one in the direction of the second is $(1,-3)$. The dot product $(1,-2)\cdot (1,-3)=7=|(1,-2)||(1,-3)|\cos \theta=5\sqrt 2 \cos \theta$. This gives $\theta=\frac 7{5\sqrt 2}\approx 0.1419$
$$d_1:y=-2x+10\Rightarrow k_1=\tan\alpha_1=-2$$ 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 $$\tan(\alpha_1-\alpha_2)=\frac{\tan\alpha_1-\tan\alpha_2}{1+\tan\alpha_1\tan\alpha_2}$$