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mathracketformulaalgebracalculus

Given 2 coordinates and an angle, find C coordinate

The following is an equilateral triangle: Bisecting ∠ACB splits the triangle into two congruent triangles (and the bisector meets the line AB at its midpoint and forms a right angle!)

if A is (1,0) and B is (5,0) and C is (c,y) and D is (c,0). C has an angle of tan(theta/2). In this case 60 degrees. How would I derive a formula in order to get the following output 3.4641016151377553.

Another example: A (-2,0), B(6,0), theta is: 120 degrees The output is: 2.309401076758504

Output must be in radians (pi/180)

Output is the C coordinate enter image description here

Solution

  • Assuming that the Points A=(a,0) and B=(b,0) are on the x-axis and that ABC is an isosceles triangle with the angle theta at C: enter image description here Then D has the coordinates ((a+b)/2,0). And in the right angled triangle ADC we have tan(theta/2) = (b-a)/2 / h. Hence we get the y-coordinate of C as h = (b-a)/2/tan(theta/2).

    Here is the corresponding racket code:

    #lang racket
(require math)
(define (half v) (/ v 2))
(define (deg2rad angle_deg) (* angle_deg (/ pi 180)))
(define (cy a b theta) (/ (half (- b a)) (tan (deg2rad (half theta)))))
(cy 1 5 60)
(cy -2 6 120)

    giving the output:

    3.464101615137755
2.309401076758504