Angle Between Two Planes in a Tetrahedron (3D Trigonometry)

Grades 11-12
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Question: Given triangle ABC with AB = 6 cm, BC = 5 cm and angle ACB = 70 degrees. DABC is a tetrahedron in which DA is perpendicular to plane ABC, and DA = 7 cm. A student claims that the angle between plane DBC and plane ABC can exceed 55 degrees. Do you agree? Explain your answer.
To determine whether the angle between plane DBC and plane ABC can exceed 55^\circ, we need to compute the exact value of this dihedral angle using geometric principles.

1. Locating the Dihedral Angle

  • Line of intersection: plane DBC and plane ABC meet along BC.
  • Construction: draw a perpendicular from point A to line BC, meeting it at H, so AH \perp BC.
  • Three Perpendiculars Theorem: since DA \perp \text{plane } ABC, DA is perpendicular to every line in that plane, so DA \perp BC. Because DA \perp BC and AH \perp BC, the Three Perpendiculars Theorem tells us the slant line DH must also be perpendicular to BC (i.e. DH \perp BC).
  • Dihedral angle: since AH \perp BC and DH \perp BC, by the definition of a dihedral angle, the angle between plane DBC and plane ABC is \angle DHA.

2. Finding the Height AH of Base Triangle ABC

In \Delta ABC, we're given AB = 6 \text{ cm}, BC = 5 \text{ cm}, and \angle ACB = 70^\circ. First, we use the Cosine Rule to find the length of side AC (call it b): \begin{align*} AB^2 &= BC^2 + AC^2 - 2(BC)(AC) \cos(\angle ACB) \\ 6^2 &= 5^2 + b^2 - 2(5)(b) \cos 70^\circ \\ 36 &= 25 + b^2 - 10b \cos 70^\circ \\ b^2 - (10 \cos 70^\circ)b - 11 &= 0 \end{align*} Substituting \cos 70^\circ \approx 0.3420: \begin{align*} b^2 - 3.420b - 11 &= 0 \\ b &= \frac{3.420 + \sqrt{3.420^2 - 4(1)(-11)}}{2} \approx 5.4417 \text{ cm} \end{align*} Next, we find the length of AH. In right triangle \Delta ACH: \begin{align*} AH &= AC \sin(\angle ACB) \\ AH &= 5.4417 \times \sin 70^\circ \approx 5.4417 \times 0.9397 \approx 5.1135 \text{ cm} \end{align*}

3. Computing the Dihedral Angle \angle DHA

In right triangle \Delta DAH (where \angle DAH = 90^\circ), with DA = 7 \text{ cm}: \begin{align*} \tan(\angle DHA) &= \frac{DA}{AH} \\ \tan(\angle DHA) &= \frac{7}{5.1135} \approx 1.3689 \\ \angle DHA &= \arctan(1.3689) \approx 53.85^\circ \end{align*}

4. Conclusion

The calculation shows the angle between plane DBC and plane ABC is about 53.85^\circ. Since 53.85^\circ < 55^\circ, the angle cannot exceed 55^\circ.
So, I disagree with the student's claim.

Diagram Key:

  • Grey plane: the base ABC.
  • Red segment DA: the tetrahedron's height, perpendicular to the base.
  • Blue point H: the foot of the perpendicular from A to BC.
  • Green angle: marks \angle DHA, the angle between the two planes.
3d-geometrytetrahedronsdihedral-angles