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I was wondering if there was a way to calculate the angle made by a line $(\space y=mx)$ in the Cartesian plane using only $m$. I used the Pythagorean theorem in this figure:Triangle $$AO= \sqrt{AB^2+OB^2}=\sqrt{x^2+m^2x^2}=x \sqrt{1+m^2}$$ Now I know that $\alpha = \cos^{-1} (\cos \alpha) $.

$$\cos \alpha = \frac{OB}{OA}=\frac{x}{x \sqrt{1+m^2}}=\frac{1}{\sqrt{1+m^2}}$$

$$\alpha = \cos^{-1} \left(\frac{1}{\sqrt{1+m^2}}\right)$$

Is this correct? Is there an easy way to solve this?

Community
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PunkZebra
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  • Use calculus and notice that $m$ is really only the derivative (the slope) of your curve at a given point, it gives you the tangent line. Therefore, simply find $\tan^{-1}\left(\frac{dy}{dx}\right)$ and you have your angle... – bjd2385 Nov 13 '14 at 19:35
  • Try Googling "inclination of a line". – John Joy Nov 14 '14 at 15:20

1 Answers1

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Looks good! Alternatively, notice that: $$ m = \frac{y}{x} = \tan \alpha $$ So we have: $$ \alpha = \tan^{-1}(m) = \tan^{-1}(0.5) \approx 26.57^\circ $$

Adriano
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