Angle between two faces of a tetrahedron $ABCD$ is a tetrahedron and $ABC$ is an equilateral triangle and $AB=BC=CA=2a$ ,given that $DA=DB=DC$ and the distance to the plane abc from $D$ is $3a$..
Find the angle between $DBC$ and $DCA$ faces
I got   $$DA=DB=DC=a  \sqrt{31/3}$$
But have no idea how to find angle between two faces..I think it should be $60^{\circ}$ between edges of tetrahedragon,but between faces...
 A: Let $BK$ be an altitude of $\Delta DCB$ and $O$ be the center of $\Delta ABC$. 
Thus, since $\Delta AKC\cong \Delta BKC$, we see that $AK\perp DC$ and we need to calculate $\measuredangle AKB$.
Now, $OC=\frac{2a}{\sqrt3}$ and since $\Delta CDO\sim \Delta ODK$, we obtain
$$\frac{OK}{OC}=\frac{DO}{DC}$$ or
$$\frac{OK}{\frac{2a}{\sqrt3}}=\frac{3a}{a\sqrt{\frac{31}{3}}},$$
which gives
$$OK=\frac{6a}{\sqrt{31}}.$$
Thus,
$$\measuredangle AKB=2\arctan\frac{AO}{KO}=2\arctan\frac{\frac{2a}{\sqrt3}}{\frac{6a}{\sqrt{31}}}=2\arctan\sqrt{\frac{31}{27}}.$$
Now, since $2\arctan\sqrt{\frac{31}{27}}>90^{\circ}$, the angle between plains $DBC$ and $DCA$ is equal to
$$180^{\circ}-2\arctan\sqrt{\frac{31}{27}}.$$
By the way, if you mean to find the measure of dihedral angle then the answer will be $$2\arctan\sqrt{\frac{31}{27}}$$
A: You can go further by doing the following steps:


*

*Compute the height $h_1$ of the face $DBC$ thrown on the side $BC$

*Compute the area of the face $DBC$ 

*Compute the height $h_2$ of the face $DBC$ thrown on the side $AD$. Denote the point on the side $AD$ appointed with this height by $E$


Now the angle we are looking for is the angle $\alpha=\angle BEC$


*

*Compute the area $P_1$ of the triangle $BEC$

*Transform the formula for the area of triangle $P=\frac{1}{2}mn\sin\alpha$ to obtain the formula for $\alpha$


 $$\alpha = \arcsin \frac{2P}{mn}$$


*Apply the new formula on the triangle $BEC$

