In a right triangle, given slope and length of hypotenuse find length of legs.

$\begingroup$

Say I have a right triangle.

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

$\endgroup$ 3

4 Answers

$\begingroup$

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

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

And

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

Two equations and two unknowns.

SPOILER ALERT:

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 $a = bm$.

$\endgroup$ 0 $\begingroup$

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 $$b = \frac{c}{\sqrt{m^2 + 1}}$$

$\endgroup$ $\begingroup$

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,

$\endgroup$ 1 $\begingroup$

$m = \frac{a}{b}$

If you take your simplified $a$ and $b$, and use $a^2+b^2=c^2$, you can do $\frac{hypotenuse}{c}$ to get the scaling factor. Once you have your scaling factor, put your slope into fraction form.

Multiply your numerator by the scaling factor and you will get $a$. Multiply your denominator by the scaling factor and you will get $b$.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy