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...
$$BM^2 = x^2 + 0.25x^2 - x^2\cos(50)$$
$$BM \approx \pm 0.78x$$
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 $\alpha - 180 \approx 79.47^\circ$.
$$\frac{BM}{\sin(50)} = \frac{x}{\sin(\alpha)}$$
$$\downarrow$$
$$\frac{0.78x}{\sin(50)} = \frac{x}{\sin(\alpha)}$$
$$\downarrow$$
$$\sin(\alpha) = \frac{x\cdot \sin(50)}{0.78x} \rightarrow \alpha \approx 79.16^\circ$$
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 $180-79.4\approx 100.6$
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.