Question Statement: Each surface of a tetrahedron ABCD is an equilateral triangle with each side 2 units long. The midpoint of AB and CD are L and M respectively. Calculate, by giving your answers correct to 3 s.f. or to the nearest 0.1 degree, a) The length of the perpendicular from A to the plane BCD b) The angle between the surface ACD and BCD c) Angle between AB and the plane BCD My Attempt So Far : I have solved part A and B. The only part that confuse me is part C. My calculation : let the angle between AB and the plane BCD be x. The perpendicular distance from A to the plane BCD be p. So sin x = p/AB = 1.63/2 But the answer given is x = 53.1 degree, which is slightly smaller than the answer I found. Is the method I used wrong?