How to plot a triangle given three side lengths?

Question

I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30). I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.

Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?

Thank you.

Answer 1

I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).

By the triangle inequality $a + b > c$. In this case you have $a + b =c$. So this will be a straight line where $C$ is in a line in between $A$ and $B$.

I tried the solution from the other topic

WHAT other topic?

and it didn't work since the result produced was C(20,0)

That is correct. $A = (0,0)$ and $B = (30, 0)$ and $C = (20,0)$ then $a = BC = \sqrt{(30-20)^2 - (0-0)^2} = \sqrt{(10)^2} = 10$ and $b = AC = \sqrt{(0-20)^2 - (0-0)^2} = \sqrt{(-20)^2} = 20$ and $c = AB = \sqrt{(0-30)^2 - (0-0)^2} = \sqrt{ (-30)^2} = 30$.

and that cant be right since one of the sides is already placed on the axis and its not a right triangle.

Who said it was a right triangle?

$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$ but $c^2 = 30^2 = 900$ and $500 \ne 900$. It isn't a right triangle.

Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?

So if $A= (0,0)$ and $B= (c,0)$ then $C = (x,y)$ where

$b^2 = AC^2 = x^2 +y^2$

$a^2 = BC^2 = (x-c)^2 + y^2$

So $a^2 - b^2 = (x-c)^2 - x^2$ or

$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$ or

$x = \frac {c^2 + b^2 -a^2}{2c}$

And so $x^2 +y^2 = b^2$ so

$y = \sqrt{ b^2 - x^2}$

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Plugging in $a = 10; b=20; c= 30$ we get

$x =\frac {c^2 + b^2 -a^2}{2c}= \frac {30^2 + 20^2 -10^2}{2*30}= \frac {900+ 400 - 100}{60} = \frac {1200}{60} = 20$

And $x =\sqrt{ b^2 - x^2}= \sqrt {20^2 - 20^2} = 0$.

So $C = (20, 0)$

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For a more proper triangle sa $a = 20; b= 15; c=25$ (an actual right triangle.

You get $A=(0,0)$ and $B = (25,0)$ and $C = (x,y)$ where

$x = \frac {c^2 + b^2 -a^2}{2c}= \frac {25^2 + 15^2-20^2}{50}= 9 $

and $y = \sqrt{ b^2 - x^2}= \sqrt{15^2 - 9^2} = 12$

So $C = (9,12)$.

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Now for an example where that fails: Say $a = 20; b=30; c = 60$. That's impossible because $a+b < c$. No such triangle exists.

If $A=(0,0)$ and $B=(60,0)$ then $C = (x,y)$ with

$x = \frac {c^2 + b^2 -a^2}{2c}= \frac {60^2 + 30^2-20^2}{60}= 10\sqrt{41}$

and $y = \sqrt{ b^2 - x^2}= \sqrt{20^2 - (10\sqrt{41}^2} = \sqrt{400-410} = \sqrt{-10}$

So $C = (10\sqrt{41},\sqrt{-10})$ is not possible.

Answer 2

Given sides a,b,and c, use the law of cosines to calculate the angle $\theta$ between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle $\theta.$