Please help me find a formula to find the 3rd point in a right triangle
5,935
Name the 90° vertex A, the 30° vertex B and the unknown vertex C
$$ \begin{pmatrix} x_C \\ y_C \end{pmatrix} = \begin{pmatrix} x_A \\ y_A \end{pmatrix} + \frac{1}{\sqrt{3}} \begin{pmatrix} (y_By_A) \\ (x_Bx_A) \end{pmatrix} $$
Example: A $=(1,1)$ B $=(5,2)$ C $= (1,1)+\frac{1}{\sqrt{3}} (1,4)$
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Javalsu
Updated on August 01, 2022Comments

Javalsu over 1 year
I'm trying to figure out how to plot a 3rd point on a graph
Given the following line segments and angles
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 xaxis.
Thanks!

Javalsu over 10 yearsThank you for the help, but it won't always be a straight line up. As the baseline won't always be parallel to the xaxiss.

Javalsu over 10 yearsThank you very much, how did you calculate the 1/√3?

John Alexiou over 10 yearsThe factor is equal to $\tan \theta$. This comes from the cosine and inner product relationship link

robkuz over 7 years@ja72 could you expand on that √3 and tan θ?

John Alexiou over 7 years$$\tan(30°) = \frac{1}{\sqrt{3}}$$ 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.