How is the area vector larger by a factor of $\frac{1}{cos \theta}$ for proving the gauss law?

So I was reading Electricity and magnetism by Purcell and I came across the Gauss law. Now in here its said I quote "The area of the patch $$A$$ is larger than that of the patch $$a$$ by two factors: first, by the ratio of the distance squared $$(\frac{R}{r})^2$$ and second, owing to its inclination, by the factor $$\frac{1}{\cos \theta}$$ "

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Now I understood the "distance squared" part. But I cant wrap my head around the inclination part.

What I tried to understand: The vector $$A$$ can be resolved into two components, one along the radial and the other perpendicular to it.

Likewise I may even resolve the radial vector into components along the vector $$A$$ and perpendicular to it.

Any help is appreciated. :)

Forget the vector $$\overrightarrow{A}$$ and imagine the surface $$A$$ itself.
It is inclined at an angle $$\theta$$ to the surface that is normal to the radius. The projected surface, orthogonal to the radius is $${{A}_{\bot }}=A\cos (\theta )$$ (If you incline the surface, it is greater !)
It is this surface $${{A}_{\bot }}$$ that you have to compare to $$a$$ a by using the factor $${{\left( R/r \right)}^{2}}$$ : $$\frac{{{A}_{\bot }}}{{{R}^{2}}}=\frac{a}{{{r}^{2}}}$$
• That does make sense. But why cant I project the radial surface onto the surface A? Yes It may be smaller than it but why cant I say that $A' \cos \theta = A$? I'm sorry but I'm not really getting on board with that idea. (btw I loved how you referred to the surface as a "she") – The Jade Emperor Feb 10 at 9:16