The equation of $AB$ is $y= 2x-6$ The equation of $AC$ is $y= -x+12$. Calculate the size of the angle $A$

Question

I have tried various methods as to answer this question using $\sin$ and $\tan$ but I cannot seem to attach my working out please send me the answers no working out needed as I just want to see where I went wrong thank you

Answer 1

$A+D+E=180°$

Now the slope of $AB$ is the same as $AD$ and it's equal to the tangent of the angle $ADE$. Instead for $AED$ we need to do $180°-AEF$ where the tangent of $AEF$ is the slope of $AE$ that is the same as $AC$.

$A+\arctan(2)+180°-\arctan(-1)=180°$

$A=\arctan(-1)-\arctan(2)$

$A=71,6°$

Answer 2

First of all calculate the slope of both these lines by comparing with the formula of a standard equation of line as : y= mx +c , c being a constant. for line AB you will find 'm1' and for line AC , you can deduce 'm2' such that 'm1' and 'm2' being the respective slopes for the lines AB and AC. Now after deducing 'm1' and 'm2', find the angle '∆' between the lines AB and AC as: tan(∆) = (m1 - m2)/(1 + m1*m2).