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Let's say I have two points $p_1=(x_1, y_1)$ and $p_2=(x_2, y_2)$, which are given as two points of a triangle $T$. And let's say I know the angles of $T$ at $p_1$ and $p_2$. How do I find the third point $p_3=(x_3, y_3)$ of $T$?

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"Find point given a line and two *angles" –  Argon May 15 '12 at 1:14
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Can't you ask the two angels? :) –  user17762 May 15 '12 at 1:14
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@Vakey Do you mean you know the angles subtended by $(x_3,y_3)$,$(x_1,y_1)$ at $(x_2,y_2)$ and $(x_3,y_3)$,$(x_2,y_2)$ at $(x_1,y_1)$? –  user17762 May 15 '12 at 1:31
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Have a look at any exposition on "angle side angle"; for example mathsisfun.com/algebra/trig-solving-asa-triangles.html –  Fixee May 15 '12 at 2:42
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@Vakey Have a look at zonalandeducation.com/mmts/intersections/… –  Fixee May 17 '12 at 16:14

1 Answer 1

up vote 2 down vote accepted

I believe that this is what you want. We can use basic linear algebra.

We have the two sides of the triangle. I write these sides as lines:

$$f(x)=m_1x+b_1=\frac{y_1-y_3}{x_1-x_3}x+b_1$$ $$g(x)=m_2x+b_2=\frac{y_2-y_3}{x_2-x_3}x+b_2$$

$m$ is the slope of the line. I assume that the angles you are talking about are the angles of elevation or depression of the lines $f$ and $g$. To turn an angle into slope, we use $\tan \theta$ where $\theta$ is the angle. Thus

$$f(x)=\tan (\theta_1)x+b_1$$ $$g(x)=\tan (\theta_2)x+b_2$$

To find $b_1$ and $b_2$, solve for them in $f(x_1)=y_1$ and $g(x_2)=y_2$ respectively.

$$f(x_1)=y_1=\tan (\theta_1)x_1+b_1\implies b_1=y_1-\tan (\theta_1)x_1$$ $$g(x_2)=y_2=\tan (\theta_2)x_2+b_2\implies b_2=y_2-\tan (\theta_2)x_2$$

To find $x_3$, we know this is the unique point such that $f(x_3)=g(x_3)$. Thus

$$ f(x_3)=g(x_3)=\tan (\theta_1)x_3+y_1-\tan (\theta_1)x_1= \tan (\theta_2)x_3+y_2-\tan (\theta_2)x_2 \implies x_3= \frac{\tan (\theta_1)x_1-\tan (\theta_2)x_2+y_2-y_1}{\tan (\theta_1)-\tan (\theta_2)}$$

I'm sure there are other ways to do this, but this is the first thing that comes to mind.

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Can you please check line functions . They should be $f(x)=\tan (\theta_1)x+b_1$ and $g(x)=\tan (\theta_2)x+b_2$. should not? –  Mathlover May 15 '12 at 12:53
    
Thanks. Though I am having trouble finding x3. I used an example where x1=0, y1=4, x2=4, y2=0, θ1 = 45, θ2 = 45. Pointing this out on a graph, I have found that x3=4 and y3=4. Using your formula above I have found that arctan(θ1)−arctan(θ2) = arctan(45)−arctan(45) = 1.5-1.5 = 0. And you can't divide by zero. Is my math wrong, or is this a special case? –  Vakey May 15 '12 at 13:27
    
@Mathlover Indeed it is $\tan$, my mistake. I never remember which one to use! –  Argon May 15 '12 at 19:48
    
@Vakey They are not both $45^\circ$. One should be $-45 ^\circ=315 ^\circ$ –  Argon May 15 '12 at 19:55
    
@Argon Thanks! That works. However, how do you get y3? –  Vakey May 16 '12 at 12:59

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