I have the following isosceles triangle:

enter image description here

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 $\alpha - 180 \approx 79.47^\circ$.

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


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


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

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.

  • 4
    $\begingroup$ The Law of Sines always gives the acute angle. When you know the angle is obtuse, you need to do the $v = 180^\circ - v$. $\endgroup$
    – Jens
    Nov 26, 2018 at 21:39

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Nov 26, 2018 at 21:42
  • $\begingroup$ I understand. Thank you. $\endgroup$
    – daedsidog
    Nov 26, 2018 at 21:43

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