Sine Rule in a 3D Pyramid: Finding a Base Angle

Grades 10-11
Student math problem: Sine Rule in a 3D Pyramid: Finding a Base Angle
Original question
Interactive Sine Rule in a 3D Pyramid: Finding a Base Angle graph — click to explore
Click the graph to explore it interactively →
Question: Explain
To solve this problem, we apply the principles of trigonometry and 3D geometry.

Part (a): Finding \angle XWY

In the base triangle WXY, we are given the lengths of two sides ( WX = 6 cm and XY = 5 cm) and an angle opposite one of them ( \angle WYX = 70^\circ). We use the Sine Rule to find \angle XWY:
\begin{align*} \frac{XY}{\sin \angle XWY} &= \frac{WX}{\sin \angle WYX} \\ \frac{5}{\sin \angle XWY} &= \frac{6}{\sin 70^\circ} \\ \sin \angle XWY &= \frac{5 \sin 70^\circ}{6} \\ \sin \angle XWY &\approx 0.78308 \\ \angle XWY &\approx 51.536^\circ \approx 51.5^\circ \end{align*}
Since WX > XY, \angle WYX must be the largest angle opposite the longer side, which confirms that \angle XWY must be acute, and there is only one valid solution.

Part (b): Finding the Angle between Triangles WXY and XYZ

1. Locating the Projection of the Apex We are given that WZ = XZ = YZ. This symmetry implies that the apex Z is equidistant from all three vertices of the base triangle. Geometrically, the projection of Z onto the plane WXY is the circumcenter of triangle WXY, which we will call point O (shown in the diagram).
2. Calculating the Height of the Pyramid In the right-angled triangle \triangle WOZ, the angle \angle ZWO is the angle between the line WZ and the base plane, which is given as 30^\circ. Let R be the circumradius WO: \begin{align*} \frac{WX}{\sin \angle WYX} &= 2R \\ \frac{6}{\sin 70^\circ} &= 2R \implies R \approx 3.1925 \text{ cm} \end{align*} The height of the pyramid ZO is: \begin{align*} ZO = R \tan 30^\circ \approx 1.8432 \text{ cm} \end{align*}
3. Determining the Dihedral Angle The angle between the triangles WXY and XYZ is the dihedral angle along their common edge XY. Let M be the midpoint of XY. Since XZ = YZ, triangle XYZ is isosceles, meaning ZM \perp XY. Similarly, in the base, O is the circumcenter, so the segment OM is the perpendicular bisector of XY, meaning OM \perp XY. Thus, the dihedral angle is \angle ZMO.
In right \triangle OMX: \begin{align*} OM = \sqrt{OX^2 - XM^2} &= \sqrt{R^2 - 2.5^2} \\ OM &= \sqrt{3.1925^2 - 2.5^2} \approx 1.9856 \text{ cm} \end{align*}
In right \triangle ZMO: \begin{align*} \tan \angle ZMO &= \frac{ZO}{OM} \approx \frac{1.8432}{1.9856} \approx 0.9283 \end{align*} Calculating the angle: \angle ZMO = \arctan(0.9283) \approx 42.87^\circ
Since 42.87^\circ < 45^\circ, the angle between the triangles does not exceed 45^\circ.

Visualization

In the interactive diagram below:
  • The blue base represents triangle WXY.
  • Point O (gold) is the circumcenter where the apex Z projects onto the base.
  • The 30° angle at W (purple) represents the inclination of the edge WZ.
  • The angle at M (red) represents the dihedral angle between the base and the back face XYZ.
  • Use the Rotation slider to view the pyramid from different perspectives.
3d-geometrytrigonometrypyramids