Solving the triangle

msdenise15

1. The problem statement, all variables and given/known data
use the information to solve the triangle: b=1, c=the square root of (3)+1, B=15 degrees.
given that sin15 degrees =the square root of 6 -the square root of 2 divided by 4. FYI.. is in the square root and than add one.

2. Relevant equations
The formulas given the law of sin sinA/a=sinB/b=sinC/c
The law cosine

3. The attempt at a solution
I used the law of sin to find angle C. The problem is written out as the square root of (3)+1 Sin(15)/1. I came out with .70 degrees for angle c. For some reason I can't get that answer again. Also I'm not sure which formula I should be using.

Gib Z

Well using the sine rule, we get [tex] \frac{\sin 15*}{1} = \frac{\sin C}{\sqrt{3}+1}[/tex].

The * I used to denote degrees. Anyway, to get the angle C, we must get Sin C by itself on one side. Once you do that, simplify the Sin 15 degrees from what it gave you in the question, and also take out a factor of sqrt2 from it. Then it should simplify very nicely, you shouldn't even need a calculator for it. You get some nice common angle, not 0.70 degrees.

HallsofIvy

msdenise15, you forgot to take the arcsine! What you calculated was that sin(C)= .707..= sqrt(2)/2. And, as Gibz said, that's a "well known" angle.