Exercise 388

In $\triangle ABC$, $\angle ABC=60^\circ$. Point $D$ is on $\overline{BC}$ so that $CD=3\cdot BD$ and $\angle DAB=30^\circ$. Find $\angle ACB$.

Solution

In $\triangle ABD$, $\angle B=60^\circ$ and $\angle DAB=30^\circ$, so $\angle ADB=90^\circ$ and $AD\perp BC$.
Let $BD=1$. Then $AD=BD\tan60^\circ=\sqrt3$ and $CD=3$.
In right triangle $ADC$, $\tan\angle ACB=\dfrac{AD}{CD}=\dfrac{\sqrt3}{3}$.
$\angle ACB=\boxed{30^\circ}$.