<s>[INST] In $\triangle ABC$, $D$ lies on $BC$, $\angle ADB=60$, $\angle ACB=45$ and $BD=2CD$. Find $\angle ABC$ The question is as the title states. In the following figure, with some given angles and two sides, the goal is to find the measure of $\angle ABC$. This problem is inspired by one which appeared in a local Math contest in Turkey. I'll post my approach to this below as an answer! Please show your own methods of solving this! 
 [/INST] Let $\angle ABC = x $, then applying the law of sines on $\triangle ADC $:
$\dfrac{AD}{\sin 45^\circ} = \dfrac{a}{\sin 15^\circ} $
Applying the law of sines on $\triangle ADB$:
$\dfrac{AD}{\sin x} = \dfrac{2 a}{\sin(120^\circ - x) } $
Dividing the two equation to eliminate $AD$ and $a$
$ \dfrac{ \sin x }{ \sin 45^\circ } = \dfrac{1}{2} \dfrac{ \sin(120^\circ - x) }{ \sin 15^\circ } $
By cross-multiplying,
$ 2 \sin x \sin 15^\circ = \sin 45^\circ \bigg( \sin 120^\circ \cos x - \cos 120^\circ \sin x \bigg) $
Dividing through by $\cos x $
$ 2 \tan x \sin 15^\circ = \sin 45^\circ \bigg( \sin 120^\circ - \cos 120^\circ \tan x \bigg)$
Hence,
$ \tan x = \dfrac{ \sin 45^\circ \sin 120^\circ }{ 2 \sin 15^\circ + \sin 45^\circ \cos 120^\circ } $
It follows that
$ x = 75^\circ $
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