h = (a / (cotB + cotC))
Can someone work out this derivation. The answer to the derivation is given above. How does one get from whatever it is to that? Angles ABC
Sides a opposite A, b opposite B, c opposite C
Altitude h (perpendicular from A to a)
Since h "cuts" a in two, I denote the two parts x and (a-x). Note that if either B or C is obtuse, either x or (a-x) is negative. This isn't that important, but it's something you should be aware of. It's acceptable to draw an acute triangle to keep things simple in your head, but recognize the possibility that h doesn't intersect a. I put x on the B side of a, and (a-x) on the C side of a. I hope this is clear.
Note that cot B = x/h, cot C = (a-x)/h. Also note that x is not a free variable - it's uniquely and exactly given by any three variables in the triangle (side lengths or angles that give congruence relations, like SSS, SAS, ASA, and so on). Solve for x:
x = h * cot B
(a-x) = h cot C
x = a - h cot C
Set the two equations equal to each other and solve for h:
h cot B = a - h cot C
h (cot B + cot C) = a
h = a/ (cot B + cot C)
Working backwards is another option, but a straightforward derivation is most rigorous.