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enter image description here 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$

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  • $\begingroup$ I think you might want to change your heading. $\endgroup$ Sep 27, 2021 at 9:00
  • $\begingroup$ Is that better? $\endgroup$ Sep 27, 2021 at 9:04
  • $\begingroup$ $152.1515593256556° $is this your answer? $\endgroup$
    – user960916
    Sep 27, 2021 at 9:16
  • $\begingroup$ Yes, if I convert the correct answer, it does result in 152. Mine results in -28. $\endgroup$ Sep 27, 2021 at 9:20
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    $\begingroup$ @ShootingStars yes. And u got your upvotes too! $\endgroup$ Sep 27, 2021 at 17:23

3 Answers 3

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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).

enter image description here

($AY$ and $PX$ are just horizontal reference lines.)

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

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  • $\begingroup$ So does arctan(-(1/e)) not directly result in the second quadrant angle??(ie do I have to add 180 to it?) $\endgroup$ Sep 27, 2021 at 12:14
  • $\begingroup$ @ShootingStars Yes, $\arctan$'s principal range (what it's able to output) is $(-90^\circ,90^\circ)$, so a negative gradient $(m)$ gives a negative acute angle $(\arctan m).$ This is actually nice and symmetrical. Just remember that angles in trigonometric functions (whether as input angles or output angles) have signs and corresponding clockwise/anticlockwise direction. Hopefully, this clarifies? $\endgroup$
    – ryang
    Sep 27, 2021 at 12:58
  • $\begingroup$ Yes, it does thanks! $\endgroup$ Sep 27, 2021 at 13:11
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$h(x) = [\ln(x)-1.5]^2-0.25$

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

$h'(e)= -1/e$ 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.

enter image description here

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

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  • $\begingroup$ Your answer is correct, and I respect that. However, you've used a formula that I have not learnt yet(I researched it though). Shouldn't my method be correct as well? I cannot figure out where I went wrong. I used simple geometry to try to solve it(ie angles in a triangle add up to 180 and supplementary angles add up to 180) but was unsuccessful. Is it possible to do it the way I did? $\endgroup$ Sep 27, 2021 at 11:49
  • $\begingroup$ @RyanG Thanks! I corrected (slope was $-e^{-1}$) $\endgroup$
    – user960916
    Sep 27, 2021 at 11:51
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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 $\approx 152^{\approx}.$

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