Trigonometry?

Question

Does the following have a solution - can it be done? No need to do it, just wondering if it can be done or not.

Given a triangle (abc) with all angles (ABC) acute. A line is drawn from each angle to perpendicular of the opposite side. (You end up with 6 overlapping right triangles). If the distances for the perpendicular lines from each angle to each side opposite are given, can you find the perimeter (a+b+c)?

Answer 1

S=sqrt{p(p-a)(p-b)(p-c)} – heron’s formula for area of triangle, where p=(a+b+c)/2; on the other hand S=a*h1/2 = b*h2/2 = c*h3/2;
Or p = S/h1+S/h2+S/h3; thus heron S^2 = S^4 (1/h1+1/h2+1/h3) (1/h1+1/h2) (1/h1+1/h3) (1/h2+1/h3), hence S^2 = 1 / {(1/h1+1/h2+1/h3) (1/h1+1/h2) (1/h1+1/h3) (1/h2+1/h3)} or S=1/sqrt(g) and a+b+c = 2S (1/h1+1/h2+1/h3) / sqrt(g) = 2 sqrt(1/h1+1/h2+1/h3) / sqrt((1/h1+1/h2) (1/h1+1/h3) (1/h2+1/h3));
m xmas!

Answer 2

Yes, and what you have at the intersection of those drawn lines is the centroid of this 2-d shape.

Answer 3

Yes. You can use the law of cosines to get the other sides.

Answer 4

yep

Answer 5

whatmetry