A triangle has one sisde with length 9 and another side with length 6. The angle opposite the side of length 6 measures 40 degrees. What is the measure of the angle opposite the side length 9?
I know we have to use law of sines.
Also in the answer you should incude how many solutions are possible. In this case I think it is 2 answers since b/asinA

Question

A triangle has one sisde with length 9 and another side with length 6. The angle opposite the side of length 6 measures 40 degrees. What is the measure of the angle opposite the side length 9?
I know we have to use law of sines.
Also in the answer you should incude how many solutions are possible. In this case I think it is 2 answers since b/asinA<1. Therefore I am pretty sure we use b1-sin^-1(b/asinA) and
 b2=180 degrees-sin^-1(b/asinA)

owlcoffee · Answer

Well, you have to set up the law of sines:
$$\frac{ a }{ \sin \angle A }=\frac{ b }{ \sin \angle B }$$
$$\frac{ 6 }{ \sin 40 }=\frac{ 9 }{ \sin \angle B }$$

owlcoffee · Answer

Remember also: 
$$\sin \alpha = \sin(180-\alpha)$$

cutiecomittee123 · Answer

I have all of that so far,plus I know SinB=9/6Sin40

anonymous · Answer

final answer is what you get with a calculator

anonymous · Answer

take the arcsine of your result

anonymous · Answer

sup cutie xd

cutiecomittee123 · Answer

@satellite73  well I plugged it into a calulator and I got 1.12 then i took the inverse sine but my  calculator says error

cutiecomittee123 · Answer

nevermind I retried it and it came out with .96, I took the inverse and I got SinB is approximately 73.74