Question

In $△ABC,DC=2BD,\angle ABC={45}^{\circ }$ and $\angle ADC={60}^{\circ }.$ Find $\angle ACB$ in degrees.
(correct answer + 3, wrong answer 0)

Answer 1

Consider a point $M$ on $AD$ such that $CM\perp AD.$ Join $B$ and $M.$
Since $\angle ADC={60}^{\circ },\angle MCD={30}^{\circ }.$
$\sin {30}^{\circ }=\frac{1}{2}$
$\Rightarrow 2MD=DC.$
$\Rightarrow BD=MD$ and $△MDB$ is isosceles.
It follows that $\angle MBD={30}^{\circ }$ and $\angle ABM={15}^{\circ }.$
We further observe that $△MBC$ is also isosceles and thus $MB=MC.$

Now, $\angle BAM=\angle BMD-\angle ABM={15}^{\circ },$ giving us yet another isosceles triangle $△BAM.$
We now have $MC=MB=MA,$ so $△AMC$ is also isosceles.
This allows us to calculate $\angle ACM={45}^{\circ }$ and finally $\angle ACB={30}^{\circ }+{45}^{\circ }={75}^{\circ }$