3D Pyramid Trigonometry Problem, Solved Step by Step

To solve this geometry problem, we will first use the Sine Rule for the base triangle and then apply 3D trigonometry to analyze the pyramid.

In triangle WXY, we are given two sides and an opposite angle. Using the Sine Rule:

Substituting the known values ( WX = 8 cm, WY = 10 cm, and \angle WXY = 50^\circ):

1. Finding the Apex Position The problem states that WZ = XZ = YZ. This geometric property implies that the vertex Z is equidistant from all vertices of the base triangle. Consequently, the projection of Z onto the base plane WXY is the circumcenter of triangle WXY. Let this point be H. The distance from H to any vertex ( W, X, Y) is the circumradius R. Using the Sine Rule again:

2. Calculating the Height ZH The angle between the edge YZ and the base WXY is given as 32^\circ. In the right-angled triangle ZHY: \tan(32^\circ) = \frac{ZH}{HY} = \frac{ZH}{R} ZH = 6.527 \times \tan(32^\circ) \approx 6.527 \times 0.6248 \approx 4.078 \text{ cm}

3. Finding the Angle between Triangles WXY and WXZ The angle between these two triangles is the dihedral angle at the edge WX. Let M be the midpoint of WX. Since XZ = WZ, triangle WXZ is isosceles, making ZM perpendicular to WX. Similarly, since H is the circumcenter, HM is the perpendicular bisector of WX. The angle we seek is \angle ZMH in the right-angled triangle ZHM. First, calculate HM: HM = \sqrt{R^2 - (\frac{WX}{2})^2} = \sqrt{6.527^2 - 4^2} \approx 5.158 \text{ cm} Now, find the angle \theta = \angle ZMH: \tan \theta = \frac{ZH}{HM} \approx \frac{4.078}{5.158} \approx 0.7906 \theta = \arctan(0.7906) \approx 38.33^\circ

Since 38.33^\circ does not exceed 40^\circ, the answer is No.

The diagram below visualizes the pyramid, showing the base triangle WXY in grey, the height ZH in red, and the dihedral angle \angle ZMH that determines the slant of face WXZ.