# Sinus-relationships in Triangle and angle above 90 degrees

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 $$\approx 111.9940863$$?

• Quick funny question. How do you make that "~" sign in Latex? Jan 27 at 17:17
• This is why it's better to use law of cosines when all sides are known. Jan 27 at 17:22
• @Vasya Definitely. Now, my question was not so much about the laws of cosines. More about why sinus gives me another answer than 111. I only ask this to get a better understanding of the sinus-relationships. Jan 27 at 17:24
• @NemanjaVuksanovic: $\sin x=a$ has two solutions for $0<x<\pi, a\ne 0, a\ne 1$: $x=\sin^{-1} a, x=\pi-\sin^{-1} a$ so you have to use additional info to pick the right solution. Jan 27 at 17:28
• My way of viewing this on is, that we have our "Unit Cirlcle", and since Sin goes from top to bottom (180 degrees), everything after those 180 degrees sin(x) start to calculate on, since the anglefrom (0,-1) and (0,1) has the same angle. Therefor, the angle from 180-sin(x) would be the same as the angle from (1,0) + sin(x). Please correct me if wrong, and please state why. Jan 27 at 17:35

My question is. Why do arcsin give us back $$\approx 68.01$$ for the angle $$B$$, when it should be $$\approx 111.9940863$$ degrees?
This is because the range of $$\arcsin(x)$$ is $$[-\pi/2,\pi/2].$$