Let the intersection of $ CE$ and $ BD$ be $ I$, the incenter. We have that the altitudes from $ I$ to $ AB$ and $ AC$ are equal. That is, $ IE\sin BEC = ID\sin CDB$ From $ BEC = 180 - C - \frac {1}{2}B$, $ CDB = 180 - B - \frac {1}{2}C$, and $ \frac {IE}{ID} = \frac {\sin 24}{\sin 18}$ (Law of Sines), we get $ \frac {\sin 24}{\sin 18} = \frac {\sin(180 - B - \frac {1}{2}C)}{\sin(180 - C - \frac {1}{2}B)} = \frac {\sin(B + \frac {1}{2}C)}{\sin(C + \frac {1}{2}B)}$ Furthermore, by analyzing triangle $ BCI$, we have $ \frac {1}{2}B + \frac {1}{2}C = 180 - BIC = 18 + 24 = 42$. It follows that both $ C + \frac {1}{2}B$ and $ B + \frac {1}{2}C$ are acute. One solution to the equations $ \frac {\sin 24}{\sin 18} = \frac {\sin(B + \frac {1}{2}C)}{\sin(C + \frac {1}{2}B)}$ $ \frac {1}{2}B + \frac {1}{2}C = 42$ is (and can be checked to be) $ B = 72,C = 12$, using $ \sin 18 = \frac {\sqrt {5} - 1}{4}$. Suppose there was another solution, and without loss of generality suppose $ B + \frac {1}{2}C > 78$ then $ C + \frac {1}{2}B$ would have to be greater than $ 12$. It follows that $ \frac {1}{2}B + \frac {1}{2}C > 42$, a contradiction. Does anybody have a solution more geometric than this? I would love to see one.

hi my ans is angleA = 96 angleB = 12 angleC = 72 (in degrees) Is it correct?

Sorry for reviving a ten year old topic.. Let $T$ be the intersection of $BD$ and $CE$. $\angle TCB + \angle TBC = 42^{\circ}$ so $\angle BAC = 96^{\circ}$ Let $M$ be the reflection of $D$ across the line $CE$ and let $N$ be the reflection of $E$ across $BD$. Since $BD$ and $CE$ are bisectors of $\angle C$ and $\angle B$, it follows that the points $M$ and $N$ are located on the line $BC$. Let the line $BD$ meet $EM$ at $R$. $\angle ERB = \angle RED + \angle EDR = 60^{\circ}$ $\angle BRN = 60^{\circ}$,$\angle MRN = 60^{\circ}$ and $\angle NDM = 24^{\circ}= \angle RDN.$ Now $N$ is the excenter of the triangle $ RDM$ and $\angle DMC= \angle RMN = \frac{\angle DMC+\angle RMN}{2}= 54^{\circ}$ So $\angle A=96^{\circ}$,$\angle B=12^{\circ}$, and $\angle C=72^{\circ}$