I have been asked to find $\angle{APB}$ in the form: $\tan^{-1}{\alpha}$ + $\tan^{-1}{\beta}$
I got the equations of the lines through differentiating the function $h(x) = {(\ln(x) – 1.5)}^{2} – 0.25$ at $e$ and ${e}^{2}$ respectively
The equation of the first line is: $$y = -xe^{-1} + 1$$
The equation of the second line is: $$y = xe^{-2} -1$$
Then after doing some math, I got:
$\angle{APB}$ = 180 – (180 – $\tan^{-1}{(\frac{-1}{e})}$ + $\tan^{-1}{(\frac{1}{e^2})}$)
$\angle{APB}$ = $\tan^{-1}{(\frac{-1}{e})}$ – $\tan^{-1}{(\frac{1}{e^2})}$
However, this is not the correct answer(due to the minus sign). I do not know if I am doing the correct work so far or if I have gone far off.
$ \angle APB$
Best Answer
Given a straight line with gradient $m,$ $$\arctan m$$ gives the acute angle measured anticlockwise from the positive $x$-direction (so, clockwise measurements are negative).
($AY$ and $PX$ are just horizontal reference lines.)
$$\measuredangle APB=\measuredangle APX+\measuredangle XPB\\=\left(180^\circ-\measuredangle YAP\right)+\measuredangle XPB\\= \left(180^\circ-(-\arctan m_1\right))+(-\arctan m_2)\\=180^\circ+\arctan\frac{-1}e-\arctan\frac1{e^2}\\=152^\circ.$$