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

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.

$$\angle APB$$

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

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.)

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

• So does arctan(-(1/e)) not directly result in the second quadrant angle??(ie do I have to add 180 to it?) Sep 27, 2021 at 12:14
• @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? Sep 27, 2021 at 12:58
• Yes, it does thanks! Sep 27, 2021 at 13:11

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

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

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

• 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? Sep 27, 2021 at 11:49
• @RyanG Thanks! I corrected (slope was $-e^{-1}$)
– user960916
Sep 27, 2021 at 11:51

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 $$\approx 152^{\approx}.$$