Using Laws of Sines for Solving SSA Triangles

[everydayimpulse]

Problem: Triangle has a=3 b=2 [alpha]=40[degrees]
So I use the Laws of Sine to find sin[beta]=0.43
and that there are two possible angles [beta],
0[degrees]<[beta]<180[degrees]
In my example is states that the two answers for [beta] are-[beta][sub-1]= 25.4[degrees]and[beta][sub-2]= 154.6[degrees]
My question is: Where/How do get the two [beta] from 0.43?? I don't know how to take 0.43 and turn it into two different degrees.

 

[tkhunny]

The inverse sine function usually is defined to produce only acute angles. You must go fishing for the obtuse companion. \(\displaystyle sin(x) = sin(\pi - x)\).

Notice that 180º - 25.4º = 154.6º