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I have a non-right triangle. I will call

the bottom or base edge $b$,
the top left edge $a$,
the top right edge $c$.

Let $ a=c+2 $ and $b=10$. How do I graph a curve where the graph $ x $ and $ y$ coordinates represent the vertex where edges $ a $ and $c$ meet?

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is there any other parameter?? –  Santosh Linkha Jun 13 '12 at 17:06
    
How do you add 2 to a line segment? Or are $a,b,c$ lengths? –  rschwieb Jun 13 '12 at 17:11
    
Yes, I'm sorry, a, b and c are line lengths. –  SubOne Jun 13 '12 at 17:22
    
Please reread the question. I made a mistake and typed a=b+2 and I meant a=c+2 –  SubOne Jun 13 '12 at 17:25

1 Answer 1

up vote 1 down vote accepted

Taking the origin as the intersection of $a$ and $b$, with the equation in terms of the angle $\theta$ at that vertex:

Through the cosine law we have:

$$(a-2)^2=a^2+100-20a\cos\theta$$ $$-4a+4=100-20a\cos\theta$$ $$a=\frac {96}{20\cos\theta-4}$$

With our particular setup, $a$ is the radius. So this is a polar equation, where

$$r(\theta)=\frac{96}{20\cos\theta-4}=\frac{24}{5\cos\theta-1}$$

we can convert to cartesian coordinates if you want:

$$5r\cos\theta-r=24$$ $$(5r\cos\theta-24)^2=r^2$$ $$(5x-24)^2=x^2+y^2$$ $$y^2=24x^2-240x+576$$ $$y^2=24(x-5)^2-24$$ $$(x-5)^2-\frac{y^2}{24}=1$$

which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\cos\theta<\frac 1 5$.

EDIT: picture time!

diagram of question setup

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I guess it's been too long. I can't comprehend this. Nor can I figure out how to graph that, or in what orientation you've placed the triangle in the graph. –  SubOne Jun 14 '12 at 15:07
    
@SubOne added a picture. hopefully it helps. I first set it up as a polar equation in $\theta$, with $a$ as the radius. Then I converted to $(x,y)$. –  Robert Mastragostino Jun 14 '12 at 16:48
    
Awsome! Thanks! –  SubOne Jun 14 '12 at 20:10

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