Angles in triangle with inscribed circle that touches one side in a ratio A triangle $\mathrm{ABC}$ has $\angle B=3\times\angle C$. The side $BC$ is devided by the touch point of the inscribed circle in the ratio 1:5. Can the angles be calculated?
With the side $a=\mathrm{BC}=6y$ I tried the "touch point" formula
$$d(B,T_a)=y=\frac{1}{2}(a+b-c)$$
combined with the cosinus theorem
$$
(6y)^2
=b^2+c^2-2bc\cos(\pi-4x)
=b^2+c^2+2bc\cos(4x)
$$
but didn't get anywhere with that. Any hints? TIA.
 A: 
Let ∠C = $x$, $r$ the radius of the incircle, and D the touch point. Then
$$\tan\frac{\angle B}2 = \tan \frac{3x}2 = \frac r{BD},\>\>\>\>\>
\tan\frac{\angle C}2=\tan \frac{x}2 = \frac r{CD}$$
Take the ratio of above equations,
$$\frac{\tan\frac{3x}2 }{\tan\frac{x}2} = \frac{CD}{BD}=5$$
Apply the identity $\tan 3a =  \frac{\tan a(3-\tan^2 a)}{1-3\tan^2 a}$ to get
$$\frac{3-\tan^2 \frac x2}{1-3\tan^2 \frac x2} = 5$$
Solve to obtain $\tan^2 \frac x2 = \frac17$, which yields $x= 2\tan^{-1}\frac1{\sqrt7}$. Thus, the angles are 
$$\angle C = x = 2\tan^{-1}\frac1{\sqrt7},\>\>\>\>\>\>\>\angle B = 3x = 6\tan^{-1}\frac1{\sqrt7}$$
A: 
Let the side $|BD|=y$, $|CD|=5y$,
and WLOG let the circumradius $R=\tfrac12$.
Then
\begin{align}
c&=\sin\gamma
,\quad
b=\sin3\gamma
,\quad
a=\sin4\gamma
,\\
y&=\tfrac12(a-b+c)
=\tfrac12\,(\sin4\gamma-\sin3\gamma+\sin\gamma)
,\\
6\,y-a&=0
,\\
2\sin4\gamma-3\sin3\gamma+3\sin\gamma&=0
,
\end{align} 
and it follows that
\begin{align}
8\cos^3\gamma-6\cos^2\gamma-4\cos\gamma+3&=0
,\\
(4\cos\gamma-3)(2\cos^2\gamma-1)
&=0
\end{align} 
with the only suitable solution
\begin{align}
\gamma&=\arccos\tfrac34
\approx 41.4^\circ
.
\end{align}
