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Say I have a right triangle.

right triangle

I know the slope and length of $c$, how do I find the length of $a$ and $b$?

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You "know the slope of $c$" in what way? Are you given an angle, or a "rise/run" fraction? – abiessu Nov 13 '13 at 22:51
Use the slope of $c$ to find the angle at $A$. – Newb Nov 13 '13 at 22:52
No need for trigonometry here. – amWhy Nov 13 '13 at 22:53
up vote 5 down vote accepted

We have a right triangle, so there are two things we know:

  • Slope $\;m = \dfrac{a - 0}{b-0}=\dfrac ab\implies a = bm$.


  • $a^2 + b^2 = \underbrace{c^2}_{\text{hypotenuse}}$

Two equations and two unknowns.


Since $a = bm, $ we can substitute $bm$ into the variable $a$ in the second equation: $$(bm)^2 + b^2 = c^2\implies b^2(m^2 + 1) = c^2 \implies b^2 = \dfrac{c^2}{m^2 + 1} \implies b = \dfrac{c}{\sqrt{m^2 + 1}}.$$ Since the lengths of the sides of a triangle must be positive, we can take the positive root of $b^2$ to solve for $b$, then back substitute to obtain $a = bm$.

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If you have the "slope" $$m = \frac ab$$ then you can write $a$ as $mb$. Fit this in $$c = \sqrt{a^2 + b^2}$$ and get $$b = \frac{c}{\sqrt{m^2 + 1}}$$

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if you kno9w the overall slope, but maybe you want to find the Y value of a shorter distance for x, use these relations.

where F is the hypotenuse, X is the x value, and Y is the Y value. Fcos(angle)=X, Fsin(angle)=Y, which you probably have memorized, but here are the ones you've forgotten.

Xtan(angle)=Y, Ycot(angle)=X,

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If you are going to discard the notation $a,b,c$ introduced in the Question, you should provide definitions for your unknowns $X,Y$. I can accept that $F$ is your hypotenuse (where the OP labels the length of hypotenuse $c$), but you have mentioned "slope" and "angle" without definition (or referring to the original quantities). – hardmath Mar 19 '14 at 4:48

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