# Right triangle, describe dx as dθ

1. Oct 2, 2009

### quantum13

1. The problem statement, all variables and given/known data
This is part of a find the electric field problem. I've narrowed down the part I find confusing.

Consider a right triangle, with legs of length x and R. Angle θ is opposite x. Leg x is a segment of a ray starting where lines x and R intersect. There is a differential length dx along line x (it is at a vertex of the right triangle). Find dx in terms of R and θ.

2. Relevant equations
standard trig functions, pythagorean theorem?

3. The attempt at a solution
The answer according to the solutions manual is dx = (R/cos^2 θ) dθ. Obviously, I cannot understand at all where that came from.

I did get R tan ( θ + dθ) = x + dx but that isn't very helpful either

2. Oct 2, 2009

### LCKurtz

Since x = R tan(θ), then dx/dθ = R sec2θ.

So what does this give for dx?

Last edited: Oct 2, 2009
3. Oct 2, 2009

### quantum13

man, this makes me feel dumb

thanks for pointing out that dx = (1 / cos2θ) dθ. i would have never guessed that trick by myself

but where does the R come into that equation?

4. Oct 2, 2009

### LCKurtz

Just a careless typo. I fixed it with an edit. dx/dθ = R sec2θ

5. Oct 2, 2009

### quantum13

this stupid part is i didnt realize it was a typo either.. but thank you for this trick. i never thought like that before