Angle chase:In $\Delta ABC, AB=AC $ and $\angle BAC=20°.$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$.

Question

In $\triangle ABC, AB=AC $ and $\angle BAC=20^\circ$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$.

Answer 1

Notice, we have $$\angle A+\angle B+\angle C=180^\circ$$ But $AB=AC\iff \angle B=\angle C$ Hence, we get $$\angle A+\angle C+\angle C=180^\circ$$ $$20^\circ+2\angle C=180^\circ$$ $$\angle C=\frac{180^\circ-20^\circ}{2}=80^\circ$$ Let the unknown angle $\angle ADC=x$

Applying sine rule in $\triangle ACD$ as follows $$\frac{\sin\angle ACD}{AD}=\frac{\sin\angle CAD}{CD}$$ Setting the corresponding values, we get $$\frac{\sin(180^\circ-(x+20^\circ))}{AD}=\frac{\sin 20^\circ}{CD}$$ $$\frac{\sin(x+20^\circ)}{AD}=\frac{\sin 20^\circ}{CD}\tag 1$$

Similarly, applying sine rule in $\triangle BCD$ as follows $$\frac{\sin\angle BCD}{BD}=\frac{\sin\angle CBD}{CD}$$ Setting $BD=AD$ & the corresponding values, we get $$\frac{\sin(x-80^\circ)}{AD}=\frac{\sin 80^\circ}{CD}\tag 2$$ Now, diving (2) by (1) we get

$$\frac{\frac{\sin(x-80^\circ)}{AD}}{\frac{\sin(x+20^\circ)}{AD}}=\frac{\frac{\sin 80^\circ}{CD}}{\frac{\sin 20^\circ}{CD}}$$ $$\frac{\sin(x-80^\circ)}{\sin(x+20^\circ)}=\frac{\sin 80^\circ}{\sin 20^\circ}$$ $$\frac{\sin x\cos 80^\circ-\cos x\sin 80^\circ}{\sin x\cos 20^\circ+\cos x\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$ $$\frac{\tan x\sin 10^\circ-\cos 10^\circ}{\tan x\cos 20^\circ+\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$ $$(\cos 20^\circ\cos 10^\circ-\sin 20^\circ\sin 10^\circ)\tan x=-2\sin 20^\circ\cos 10^\circ$$ $$\cos 30^\circ\tan x=-2\sin 20^\circ\cos 10^\circ$$

$$\tan x=\frac{-2\sin 20^\circ\cos 10^\circ}{\cos 30^\circ}=\frac{-2\sin 20^\circ\cos 10^\circ}{\frac{\sqrt 3}{2}}$$ $$\tan x=\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}$$$$\iff x=\tan^{-1}\left(\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$ $$=180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$

Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\angle ADC}=\color{blue}{180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)\approx 142.12^\circ}}$$

Answer 2

ADC = t , ACD = x

AB = AC , AD = BD

In ACD : AC / sint = AD / sin(80-x)

In ABC : AC / sin 80 = BC / sin 20

In BDC : AD / sin 80 = BC / sin(180-x)

x = 60 , t = ADC = 140