Why do two applications of the law of sines give different angles?

[Aidyan]

Homework Statement

Given is a triangle with sides a=3.1cm, b=5cm, c=4.7cm and opposite angle to side a, α=36°. I must find out for angles β and γ using the law of sines.

Homework Equations

Law of sines.

The Attempt at a Solution

[/B]
I first tried:
sin(γ)}/c=sin(α)/a ⇒ γ≈63.02°
β=180-α-γ=80.98°

However, one could also calculate:
sin(β)}/b=sin(α)/a ⇒ β≈71.45°
γ=180°-α-β=72.55°

Two approaches, that I believed to be equivalent, but which furnish different results. So, I obviously make something wrong. But ... I don't get it... what is it?

 

[phyzguy]

I think you have applied the law of sines correctly. The problem is that the triangle you have given is impossible. Try applying the law of cosines to the values you have given.

 

[Aidyan]

I draw the triangle on a piece of paper. What is wrong with this drawing?

 

[phyzguy]

Try measuring side a. If you look at it, side is is more like 3.0-3.1, not 3.7. This is also what the law of cosines would tell you.

Edit: I see now that I misread your 3.1 as a 3.7. If you accept your values of 36 degrees, 5.0 and 4.7, then a must be 3.011. If you use this value, it all works out.

 

[phyzguy]

Note that once you have specified b, c, and alpha, this fixes the shape of the triangle. The other side and the other two angles are then determined. You can't make them whatever you want.

 

[Aidyan]

Ahhh... now I see that. Of course, I first draw the triangle, then took the measures of all three sides and checked... but obviously that couldn't work that way. Thanks so much!