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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) the question

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$$d_1:y=-2x+10\Rightarrow k_1=\tan\alpha_1=-2$$ 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 $$\tan(\alpha_1-\alpha_2)=\frac{\tan\alpha_1-\tan\alpha_2}{1+\tan\alpha_1\tan\alpha_2}$$

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  • $\begingroup$ Explanation? ${}{}{}$ $\endgroup$
    – Pedro Tamaroff
    Apr 24 '13 at 21:15
  • $\begingroup$ how was ∠(d1,d2)=arctan(k1−k2)/(1+k1k) derived ?? $\endgroup$ Apr 25 '13 at 9:53
  • $\begingroup$ pardon me for not getting it but by arctan do you mean tangent inverse or just tangent (in radians) $\endgroup$ Apr 25 '13 at 10:01
  • $\begingroup$ Edited and explained $\endgroup$
    – Adi Dani
    Apr 25 '13 at 10:24
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Try using this relationship. If a line has slope $m$, and the angle at which it hits the $x$-axis is $\theta$, then $$\tan(\theta) = m.$$

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  • $\begingroup$ The relationship shown really did the thing thanks :D $\endgroup$ Apr 24 '13 at 19:45
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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 $\theta=\frac 7{5\sqrt 2}\approx 0.1419$

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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.

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    $\begingroup$ thanks your suggestion and @ncmathsadist given relation really helped :D $\endgroup$ Apr 24 '13 at 19:42

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