# Using the Law of Cosines results in an "invalid" answer, why?

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.

• The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^\circ - v$.
– Jens
Nov 26, 2018 at 21:39

$$\sin \alpha = \frac {\sin 50^\circ}{\sqrt {1.25 - \cos 50^\circ}}$$
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.
• Just to drive the point home to the OP: discarding the negative answer is correct. The law of cosines works just fine, but when using $\arcsin$ to solve problems it only finds one possible solution, not all possible solutions. Nov 26, 2018 at 21:42