Point $D$ lies on side $BC$ of triangle $ABC$. $DC = 2BD$. Given that $\angle ABC = 45^{\circ}$ and $\angle ADC = 60^{\circ}$, find the remaining angles of triangle $ABC$.

1. Problem Description

Point $D$ lies on side $BC$ of triangle $ABC$. $DC = 2BD$. Given that $\angle ABC = 45^{\circ}$ and $\angle ADC = 60^{\circ}$, find the remaining angles of triangle $ABC$.

2. Solution Steps

Let $\angle BAC = \alpha$ and $\angle ACB = \gamma$. Also let $BD = x$, which implies $DC = 2x$, and thus $BC = 3x$.

In $\triangle ABD$, we have $\angle ADB = 180^{\circ} - \angle ADC = 180^{\circ} - 60^{\circ} = 120^{\circ}$. Then $\angle BAD = 180^{\circ} - \angle ABD - \angle ADB = 180^{\circ} - 45^{\circ} - 120^{\circ} = 15^{\circ}$.

In $\triangle ADC$, we have $\angle DAC = 180^{\circ} - \angle ADC - \angle ACD = 180^{\circ} - 60^{\circ} - \gamma = 120^{\circ} - \gamma$.

Since $\angle BAC = \angle BAD + \angle DAC$, we have $\alpha = 15^{\circ} + 120^{\circ} - \gamma = 135^{\circ} - \gamma$.

Also, since the sum of the angles in $\triangle ABC$ is $180^{\circ}$, we have $\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.

Therefore, $\alpha + 45^{\circ} + \gamma = 180^{\circ}$. Substituting $\alpha = 135^{\circ} - \gamma$, we get $135^{\circ} - \gamma + 45^{\circ} + \gamma = 180^{\circ}$.

The above equation simplifies to $180^{\circ} = 180^{\circ}$, which doesn't help us find $\gamma$.

Applying the Law of Sines in $\triangle ABD$,

$\frac{BD}{\sin(\angle BAD)} = \frac{AD}{\sin(\angle ABD)}$

$\frac{x}{\sin(15^{\circ})} = \frac{AD}{\sin(45^{\circ})}$

$AD = \frac{x \sin(45^{\circ})}{\sin(15^{\circ})}$

Applying the Law of Sines in $\triangle ADC$,

$\frac{DC}{\sin(\angle DAC)} = \frac{AD}{\sin(\angle ACD)}$

$\frac{2x}{\sin(120^{\circ} - \gamma)} = \frac{AD}{\sin(\gamma)}$

$AD = \frac{2x \sin(\gamma)}{\sin(120^{\circ} - \gamma)}$

Therefore, $\frac{x \sin(45^{\circ})}{\sin(15^{\circ})} = \frac{2x \sin(\gamma)}{\sin(120^{\circ} - \gamma)}$.

$\frac{\sin(45^{\circ})}{\sin(15^{\circ})} = \frac{2 \sin(\gamma)}{\sin(120^{\circ} - \gamma)}$

$\sin(120^{\circ} - \gamma) \sin(45^{\circ}) = 2 \sin(\gamma) \sin(15^{\circ})$

Since $\sin(15^{\circ}) = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$,

$\sin(120^{\circ} - \gamma) \frac{\sqrt{2}}{2} = 2 \sin(\gamma) \frac{\sqrt{6}-\sqrt{2}}{4}$

$\sin(120^{\circ} - \gamma) \frac{\sqrt{2}}{2} = \sin(\gamma) \frac{\sqrt{6}-\sqrt{2}}{2}$

$\sin(120^{\circ} - \gamma) \sqrt{2} = \sin(\gamma) (\sqrt{6}-\sqrt{2})$

$\sin(120^{\circ} - \gamma) = \sin(\gamma) (\sqrt{3}-1)$

$\sin(120^{\circ}) \cos(\gamma) - \cos(120^{\circ}) \sin(\gamma) = \sin(\gamma) (\sqrt{3}-1)$

$\frac{\sqrt{3}}{2} \cos(\gamma) + \frac{1}{2} \sin(\gamma) = \sin(\gamma) (\sqrt{3}-1)$

$\sqrt{3} \cos(\gamma) + \sin(\gamma) = 2 \sin(\gamma) (\sqrt{3}-1)$

$\sqrt{3} \cos(\gamma) = \sin(\gamma) (2\sqrt{3}-2-1) = \sin(\gamma) (2\sqrt{3}-3)$

$\tan(\gamma) = \frac{\sqrt{3}}{2\sqrt{3}-3} = \frac{\sqrt{3}}{2\sqrt{3}-3} \cdot \frac{2\sqrt{3}+3}{2\sqrt{3}+3} = \frac{6+3\sqrt{3}}{12-9} = \frac{6+3\sqrt{3}}{3} = 2+\sqrt{3}$

$\gamma = 75^{\circ}$

$\alpha = 135^{\circ} - \gamma = 135^{\circ} - 75^{\circ} = 60^{\circ}$

3. Final Answer

$\angle BAC = 60^{\circ}$, $\angle ACB = 75^{\circ}$