The angle $ \angle APB$ between the tangents to a curve.

Question

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$

Answer 1

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.)

$$\measuredangle APB=\measuredangle APX+\measuredangle XPB\\=\left(180^\circ-\measuredangle YAP\right)+\measuredangle XPB\\= \left(180^\circ-(-\arctan m_1\right))+(-\arctan m_2)\\=180^\circ+\arctan\frac{-1}e-\arctan\frac1{e^2}\\=152^\circ.$$

Answer 2

$h(x) = [\ln(x)-1.5]^2-0.25$

$h'(x) = \frac{2(\ln(x)-1.5)}{x}$

$h'(e)= -1/e$ and $h'(e^2) = \frac{1}{e^2}$

$\tan(\theta) =$ $\tan^{-1}$|$\frac{m1-m2}{1+m1.m2}$|else it's supplementary

Formula : $\tan^{-1}${|$\frac{m1-m2}{1+m1.m2}$|}

$\theta' = 180 -\theta =152.1515593256556 $

• As per the question diagram or using geometry.

Here, At point A angle using derivative will be negative $(\tan^{-1}(-e^{-1}))$ which is negative.

Answer 3

In any triangle external angle made by producing a line of any triangle equals the sum of two opposite angles.

$$ \gamma_2- \gamma_1 $$

made at x-coordinate locations $(e^2,e) $ respectively

As you marked, adopting a consist anti clockwise rotation convention reckoned positive

$$ \pi+\tan^{-1} \alpha - (\pi-\tan^{-1} \beta )$$

$$ \tan^{-1} \alpha + \tan^{-1} \beta $$

Now $ \beta <0,$ then only can you get arctan obtuse between $(\pi/2,\pi) $ in second quadrant.

Numerical calculation results in a value $\approx 152^{\approx}.$