Please help me find a formula to find the 3rd point in a right triangle

geometry trigonometry

5,935

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

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

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

Alhpa

Wolfram Alpha Link

5,935

Javalsu

Updated on August 01, 2022

Comments

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 x-axis.

Thanks!
Javalsu over 10 years

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 over 10 years

Thank you very much, how did you calculate the 1/√3?
John Alexiou over 10 years

The 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.