Angle between 2 faces of tetrahedron

Two faces ABC and DBC of a tetrahedron ABCD are right-angled triangles with $\angle ACB = \angle DCB = 90^{\circ}$.

Given that the edge DA is perpendicular to the face $ABC$, $\angle CBD = 45^{\circ}$ and the angle between the line $DB$ and the face $ABC$ is $30^{\circ}$.

Find the angle between the two faces $ABC$ and $DBC$.

I am not sure how to go about this problem. I made a paper tetrahedron to try and visualize this problem better, and what I get is that the angle should be $45^{\circ}$. I am however not able to make the connection on how to go about proving this.

Can you guys please help me figure this out? Thanks again for all your help.

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2 Answers

Put $C$ at the origin, $B$ on the $x$-axis, and $A$ on the $y$-axis; $D$ is in the $yz$-plane, $\overline{DA}$ is perpendicular to the $xy$-plane, $\overline{DB}$ makes a $30$º degree angle with the $xy$-plane (so that $\angle DBA = 30$º), $\angle CBD = 45$º, and what’s wanted is $\angle DCA$.

$\triangle DCB$ is an isosceles right triangle; there is no harm in assuming that $|CB| = 1$, in which case $|CD| = 1$, and $|DB| = \sqrt 2$. $\angle BAD$ is a right angle, since $\overline{DA}$ is perpendicular to the $xy$-plane, and $\angle DBA = 30$º, so $\triangle DBA$ is a $30-60-90$ right triangle, and $|DA| = \frac{|DB|}{2} = \frac{\sqrt 2}{2}$. Finally, $\triangle DCA$ is a right triangle with hypotenuse $|DC| = 1$ and leg $|DA| = \frac{\sqrt 2}{2}$, so the remaining side is $\sqrt{1^2 - \left( \frac{\sqrt 2}{2} \right)^2}= \frac{\sqrt 2}{2}$, $\triangle DCA$ is an isosceles right triangle, and $\angle DCA$ is indeed $45$º.

This is a little more than I should probably say for homework, but I’ve deliberately been a bit concise, and you’re still going to have to visualize it properly to follow the calculations.

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Can you please explain why $\triangle DCA$ is a right triangle? Thanks for your help, the fog is starting to clear up! –  mathguy80 Jul 13 '11 at 6:59
Aha! got it, $DA \perp face ABC$, hence by definition $DA \perp AC$, hence... –  mathguy80 Jul 13 '11 at 7:01
There you go! Always nicer when you can do it yourself, isn’t it? –  Brian M. Scott Jul 13 '11 at 7:19
Yup! Your solution is pretty much complete, but it was good to follow the correct line of reasoning. Thanks again. –  mathguy80 Jul 15 '11 at 11:10

If the angle between $DB$ and face $ABC$ is $30^\circ$, then (by definition) the angle between $DB$ and a line perpendicular to $ABC$ is $60^\circ$; draw in such a perpendicular: namely, let $P$ be the point on $AC$ such that $DP \perp AC$ (and thus $DP \perp ABC$, so that also $BP \perp DP$).

Use $\triangle BCD$ and then $\triangle BDP$ to determine the lengths of segments $BD$ and $DP$ in turn; then you should be able to determine $\angle ACD$, which (because $BD \perp \triangle ACD$) is equal to the desired dihedral angle.

(Hmmmm ... I didn't seem to need the fact that $DA \perp \triangle BCD$.)

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Thanks for your help. I'll try to work it out with this method and get back later. –  mathguy80 Jul 13 '11 at 13:39
You are right, the 30-60-90 triangle can be implied. Regarding the $\perp$ wouldn't point $C$ be same as $P$, since they are right-triangles? –  mathguy80 Jul 15 '11 at 11:07