Angle between two faces of a tetrahedron

Question

$ABCD$ 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...

Answer 1

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 $BEC$

Answer 2

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 $$2\arctan\sqrt{\frac{31}{27}}$$