Assume we have the following triangle $A$, $B$ and $C$, where we only know $A$ and all the sides of the triangle. We have the following now.
$A = 25 \text{ degrees}$
$B = ?$
$C = ?$
$a = 4.33$
$b = 9.5$
$c = 7$
(I also posted a picture of the triangle. This is only for illustration purposes)
I now want to find the angles for B and C, for that, I use the sinus-relationships, which is defined as the following…
$$\frac{\sin(a)}{a} = \frac{\sin(b)}{b} = \frac{\sin(c)}{c}$$
If you start by calculating the angle for B, you get $$\frac{\sin(a)}{a} = \frac{\sin(b)}{b}$$
$$\frac{\sin(a)}{a}*b = \sin(b)$$
$$\frac{\sin(25)}{4.33}*9.5 = \sin(b)$$
$$\frac{\sin(25)*9.5}{4.33} = \sin(b)$$
$$\sin(b) \approx 0.9272225142$$
$$\arcsin(b) \approx 68.01$$
Doing the last angle I get $\approx 43.10$ degrees.
My question is. Why do arcsin give us back $\approx 68.01$ for the angle $B$, when it should be $\approx 111.9940863$ degrees? I know you can take the angle and subtract it from $180$ to get the actual degree, however, what is it that makes arcsin give back $\approx 68.01$ instead of $\approx 111.9940863$?
Best Answer
This is because the range of $\arcsin(x)$ is $[-\pi/2,\pi/2].$