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I'm trying to figure out how to plot a 3rd point on a graph

Given the following line segments and angles

enter image description here

Is there a formula for the 3rd point?

Note: This image is just for an example. The base line of the triangle will not always be parallel to the x-axis.


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up vote 2 down vote accepted

Name the 90° vertex A, the 30° vertex B and the unknown vertex C

$$ \begin{pmatrix} x_C \\ y_C \end{pmatrix} = \begin{pmatrix} x_A \\ y_A \end{pmatrix} + \frac{1}{\sqrt{3}} \begin{pmatrix} -(y_B-y_A) \\ (x_B-x_A) \end{pmatrix} $$

Example: A $=(1,1)$ B $=(5,2)$ C $= (1,1)+\frac{1}{\sqrt{3}} (-1,4)$


Wolfram Alpha Link

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Thank you very much, how did you calculate the 1/√3? – Javalsu May 6 '13 at 19:57
The factor is equal to $\tan \theta$. This comes from the cosine and inner product relationship link – ja72 May 6 '13 at 23:18
@ja72 could you expand on that √3 and tan θ? – robkuz May 3 at 18:46
$$\tan(30°) = \frac{1}{\sqrt{3}}$$ Also for a vector $(x,y)$ the perpendicular is $(-y,x)$ so point C is on the line perpendicular to AB and passing through A. – ja72 May 3 at 21:07

There are various ways of finding it. The one I would favour in this particular scenario is:

  1. Work out the length of the edge from the right angle to the point, using trig.
  2. Find a unit vector from the right angle to the point.
  3. Multiply and add.

The interesting one in the non-axis-aligned case is point 2, but it has a simple trick. You can easily find a unit vector from the right angle to the other point (vector subtraction), and then a 90 degree rotation takes $(x, y)$ to $(\pm y, \mp x)$ with the sign depending on the direction of rotation.

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Seeing as it is a right angled triangle, we know that the angle at the point $(?,?)$ is going to be $180 - (90 + 30) = 60$. Using the sine rule, we can then work out the length of that vertical line, which we get to be:

$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$$ $$\frac{a}{\sin(30)} = \frac{10}{\sin(60)} \implies a = \frac{10}{\sin(60)} \cdot \sin(30) = \frac{10}{\sqrt3}.$$

This is your $y$ co-ordinate, and as you know that it is in a straight line up, you're $x$ co-ordinate will remain the same and so your point will be

$$\left( 10, \frac{10}{\sqrt3} \right).$$

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Thank you for the help, but it won't always be a straight line up. As the baseline won't always be parallel to the x-axiss. – Javalsu May 6 '13 at 18:34

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