A triangle is given with sides $a = 6$, $b = 7$, $c = 9$; required to find the angles of the triangle. The solution I'm familiar with is to find the angles using the half-angle formulas for triangles:
$\tan$$A\over 2$$ = \sqrt{(s–b)(s–c)\over s(s–a)}$, where $s$ is the semiperimeter
By thus formula, $\tan$$A\over 2$$= 0.74$ (approx.)
Hence $A = 73°$ (approx.)
Solving for the rest of the angles using the same formula,
$B = 77°$ (approx.) $C = 79°$ (approx.)
However, these angles obviously don't add up to $180°$. What am I doing wrong? Thank you in advance.