What is a symbolic expression to calculate height of a triangle from two angles and a side?

Question

Given one side (a), which we'll call the 'base', and it's adjacent angles (B, C), calculate the 'height' of a triangle. In this case 'height' means the distance from the opposite vertex of the given side to a line co-linear to the given side.( I.e., this is an Angle-Side-Angle (ASA) problem).

It's a simple enough problem I thought solutions would be all over the web. I was surprised after 20+ min of google to not find the answer, however. Everybody who explains this problem (including on stack-exchange) does so by inserting actual numbers throughout the calculation. Just to re-iterate, this question is not about how to solve the problem, but what the symbolic function representing the solution is.

In other words:

h(a, B, C) = ?

Answer 1

If we knew $b$, then the area of the triangle would be $$ \mathrm{Area} = \frac{1}{2}ab\sin C. $$ But we don't know $b$. We do know the law of sines, though, which says $b/\sin B = a/\sin A$. Since $A + B + C = \pi$, $\sin A = \sin(\pi - B - C) = \sin(B + C)$. So $b = a \sin B/\sin(B+C)$, and from that we get that the area of the triangle is $$ \mathrm{Area} = \frac{a^2}{2}\frac{\sin B\, \sin C}{\sin(B+C)} $$ Since $\mathrm{Area} = ah/2$, this gives $$ h= a\frac{\sin B \,\sin C}{\sin(B+C)} = \frac{a}{\cot \, B + \cot \, C}, $$ where the second equality follows from $\sin(B+C) = \sin B \,\cos C + \cos B\,\sin C$.

Answer 2

why won't you use the Pythagorean Theorem? Let $D$ be the base Point of the hight $h$ then $$h^2+(a+x)^2=c^2$$ where $x=CD$ and $$\sin(180^{\circ}-\gamma)=\frac{h}{x}$$ $$\frac{\sin(180^{\circ}-\beta-\gamma)}{\sin(\gamma)}=\frac{a}{c}$$ this gives a formula for $h$ $$\left(\frac{h}{\sin(\gamma)}+a\right)^2+h^2=\frac{a^2\cdot \sin^2(\gamma)}{\sin^2(\beta+\gamma)}$$ and we get for $h$ $$h=\frac{-2a\csc(\gamma)-\sqrt{4\csc^2(\beta+\gamma)a^2+4\csc^2(\beta+\gamma)\sin^2(\gamma)a^2-4a^2}}{2(\csc^2(\gamma)+1)}$$

Answer 3

I went ahead and worked it out myself, and I get the following. Somebody please correct me if this is wrong. I believe this is general - even if one of the angles is obtuse (see the bottom triangle in image attached to the question).

$$ h(a,B,C) ~=~ \frac{a}{\cot{B}+\cot{C}} ~=~ a ~\frac{\tan{B}\tan{C}}{\tan{B}+\tan{C}} $$