Dear Uncle Colin,

I need to find an angle! ABC is a triangle with median AD, while angles BAD and CAD are 110º and 20º, respectively. What’s angle ACB?

-- Angle Being Evasive, LOL

Hi, ABEL, and thanks for your question! Even if you’ve used degrees. For heaven’s sake, get a grip.

I’d start by drawing a picture, which you can do yourself ((Honestly, do you expect me to do everything?)), and asking what I can make concrete. If I call the missing angle ACB $\theta$, then I know that CDA is $160-\theta$, CDB is $20+\theta$ and ABC is $50+\theta$. Not all of these will prove useful, but they’re good to have.

We also know that the lengths CD and DB are the same, because AD is a median. Let’s call that $x$ and the lenght of AB $y$.

I’m going to work with two triangles separately, applying the sine rule to both ACD and ADB.

For ACD, I have $\frac{y}{\sin(\theta)}= \frac{x}{\sin(20)}$. For ADB, I’ve got $\frac{x}{\sin(110)} = \frac{y}{\sin(50-\theta)}$.

I neither know nor care about $x$ and $y$, and I’m going to eliminate them. In the first triangle, $\frac{y}{x} = \frac{\sin(\theta)}{\sin(20)}$ and in the second, $\frac{y}{x} = \frac{\sin(50-\theta)}{\sin(110)}$.

There’s also a nice symmetry: $\sin(110) = \sin(70) = \cos(20)$, which we can use to good effect.

Doing that substitution, we have $\frac{\sin(\theta)}{\sin(20)} = \frac{\sin(50-\theta)}{\cos(20)}$. Rearranging, we get $\frac{\sin(\theta)}{\sin(50-\theta)} = \tan(20)$, which I’ll call $T$.

Now, $\sin(\theta) = T \sin(50-\theta)$. Expanding the right hand side gives $\sin(\theta) = T \sin(50)\cos(\theta) - T \cos(50)\sin(\theta)$.

Ah, all of those lovely $\theta$s! You know, we can divide them all by $\cos(\theta)$ and make it even simpler: $\tan(\theta) = T \sin(50) - T \cos(50) \tan(\theta)$


$\tan(\theta) \left[ 1 + T \cos(50) \right] = T\sin(50)$

$\tan(\theta) = \frac{\tan(20)\sin(50)}{1 + \tan(20)\cos(50)}$

Most reasonable people would incur the Mathematical Ninja’s displeasure at that point by reaching for a calculator; even a cursory “that’s about $\frac{\frac{1}{3}\frac{4}{5}}{1 + \frac{1}{3}\frac{3}{5}} = \frac{\frac{4}{15}}{\frac{6}{5}} = \frac{4}{18}$, about 0.222” would do him (it’s actually 0.226, so well done us), and the arctangent of that is in the 12-13 degree region. Or, if you want to do it the easy way, 12.73.

Hope that helps!

-- Uncle Colin