Math HW question

In triangle ABC, b = 20,c = 12, m angle A = 23 degrees 50 minutes. Find "a" to the nearest unit.

cos-rule:

$\begin{array}{rcll} a^2 &=& b^2+c^2-2bc\cos{( A )} \qquad & b = 20\ \rm{m} \qquad c = 12\ \rm{m} \qquad A =23^{\circ}\ 50' \\\\ a^2 &=& 20^2+12^2-2\cdot 20 \cdot 12 \cdot \cos{( 23^{\circ}\ 50' )} \\ a^2 &=& 400+144-480 \cdot \cos{( 23^{\circ}\ 50' )} \\ a^2 &=& 544-480 \cdot \cos{( 23^{\circ}\ 50' )} \qquad & 23^{\circ}\ 50' = 23+\frac{50}{60}=23.8\overline{3}^{\circ}\\ a^2 &=& 544-480 \cdot \cos{( 23.8\overline{3}^{\circ} )} \\ a^2 &=& 544-480 \cdot 0.91472473989 \\ a^2 &=& 544-439.067875145 \\ a^2 &=& 104.932124855 \qquad & \sqrt{}\\ a &=& 10.2436382626 \\ \end{array}$

a = 10 m

I think its 66 degrees and 40 minutes

WHAT KIND OF A TRIANGLE IS IT? RIGHT-ANGLE, SCALENE, ISOSCELES.......ETC.