Trigonometry solving triangles

Question

Consider a triangle which has side lengths a, b, and c, and angles , , and , with  opposite the side of length a, and so on.

Suppose   ,   , and   . How many possible solutions are there?

Answer 1

Using the law of sines,

sin (π/6) / 3 = sin α / 7, which gives

sin α = 7/6 > 1, so there are no solutions

Answer 2

If we use the law of cosines starting with "b", we can solve for possibilities for side "c":

b² = a² + c² - 2ac•cos(β)

3² = 7² + c² - 2•3•c•cos(π/6)

9 = 49 + c² - 3√3c

c² - 3√3c + 40 = 0

c = [3√3 ± √((-3√3)2 - 4(1)(40))]/(2(1))

c = (3√3 ± √(27 - 160))/2 = (3√3 ± √-133)/2 so we have imaginary solutions. This means there would be no triangle possible.