Angles created by triangle median

Question

Given that $AD$ bisects the line $BC$ into equals parts $BD$ and $DC$, does that tell us anything about the angles $x$ and $y$?

We also know $\angle B$ and $\angle C$. Is all this enough info to find angles $x$ and $y$ ?

Answer 1

From the law of sines

$ \dfrac{ AD }{\sin 20^\circ } = \dfrac{ DC }{\sin y } $

and

$ \dfrac{ AD}{\sin 40^\circ } = \dfrac{ BD }{\sin x } $

Since $ DC = BD $ , then dividing out the two equations gives

$ \dfrac{\sin 40^\circ}{\sin 20^\circ} = \dfrac{\sin x}{\sin y } $

In addition to this we know that $ x + y = 180^\circ - 20^\circ - 40^\circ = 120^\circ $

Hence, we now have

$ \sin (120^\circ - y) \sin 20^\circ = \sin 40^\circ \sin y $

Expanding the left hand side using compound angle identity,

$ (\sin 120^\circ \cos y - \cos 120^\circ \sin y ) \sin 20^\circ = \sin 40^\circ \sin y $

Divide through by $\cos y$

$ (\sin 120^\circ - \cos 120^\circ \tan y ) \sin 20^\circ = \sin 40^\circ \tan y $

So that

$ \tan y = \dfrac{ \sin 120^\circ \sin 20^\circ }{\sin 40^\circ + \sin 20^\circ \cos 120^\circ } $

From which you can calculate $y$. Angle $x$ follows, as $x = 120^\circ - y $.