Btukfyl

I have the following isosceles triangle:

I want to find the $alpha$ angle, and I know that it is obtuse.

My first instinct was to get the length of $BM$ using the Law of Cosines, which results in two answers: a negative one and a positive one; I immediately descredited the negative one because all lengths are assumed to be positive in geometry, or so have I assumed thus far...

$$BM^2 = x^2 + 0.25x^2 - x^2cos(50)$$

$$BM approx pm 0.78x$$

From here, I thought I could easily extrapolate $alpha$ by plugging it into the Law of Sines formula, but to my surprise I did not get the correct result, $alpha approx 100.53^circ$, but $alpha - 180 approx 79.47^circ$.

$$frac{BM}{sin(50)} = frac{x}{sin(alpha)}$$

$$downarrow$$

$$frac{0.78x}{sin(50)} = frac{x}{sin(alpha)}$$

$$downarrow$$

$$sin(alpha) = frac{xcdot sin(50)}{0.78x} rightarrow alpha approx 79.16^circ$$

I assume this is because I discredited what is a valid trigonometrical answer, but why is it? Until now I have been under the impression that all lengths of geometrical shapes must be positive.

I am aware there are other methods to solve this, but I am only particularly interested in why my specific one does not behave the way I want it to.

Thanks in advance.

You have:

$sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$

There are 2 values for $alpha$ between $0$ and $180^circ$ such that $sin alpha = frac {sin 50^circ}{sqrt {1.25 - cos 50^circ}}$

one is approximately $79.4^circ$ the other is $180-79.4approx 100.6$

The $arcsin$ function on your calculator will return an answer in the interval $[-90,90]$ and you may need to go from there to find the angle you are actually looking for.