As shown in the figure, in the triangular pyramid p-abc, PA = Pb = PC, if PA ⊥ Pb, PA ⊥ PC, Pb ⊥ PC, find the angle between PA and plane ABC

As shown in the figure, in the triangular pyramid p-abc, PA = Pb = PC, if PA ⊥ Pb, PA ⊥ PC, Pb ⊥ PC, find the angle between PA and plane ABC

PA ⊥ Pb, PA ⊥ PC, Pb ⊥ PC, PA = Pb = PC ⊥ AB = BC = AC, side PAB, PAC, PBC are isosceles triangle, bottom ABC is equilateral triangle, take the midpoint m of BC, connect PM, am ⊥ am ⊥ BC, PM ⊥ BC ⊥ BC ⊥ plane PAM ⊥ BC ⊥ bottom PAM ⊥ bottom ABC ⊥ Ma is the projection of PA in bottom ABC

In the triangular pyramid p-abc, PA ⊥ Pb, Pb ⊥ PC, PA ⊥ PC, O are the photography of P on plane ABC. It is proved that O is the perpendicular center of △ ABC

It is proved that: if ∵ Pb ⊥ PA, Pb ⊥ PC  & nbsp; Pb ⊥ plane PAC  & nbsp; Pb ⊥ AC & nbsp; & nbsp; & nbsp; & nbsp; and ∵ o is the projection of P on plane ABC, then Po ⊥ plane ABC ∵ & nbsp; PO ⊥ AC & nbsp; & nbsp; & nbsp; & nbsp; ② AC ⊥ plane Pb ≁ AC ⊥ Bo can prove & nbsp; AO ⊥ BC ≁ by the same theorem

Let p be any point in the equilateral triangle ABC

Because PA 〈 AB is pa 〈 BC
And Pb + PC 〉 BC (the sum of the two sides of a triangle is greater than the third side)
So PA 〈 BC 〈 Pb + PC
That is pa 〈 Pb + PC

Let two independent three event ABC satisfy the following conditions: a ∩ B ∩ C = empty set, PA = Pb = PC ＜ 1 / 2, and P (a ∪ B ∪ C) = 9 / 16. Let P (a) = 1/
Thank you

P(A∪B∪C) =PA+PB+PC-PAB-PAC-PBC+PABC;
So PAB = PA * Pb PBC = Pb * PC PAC = PA * PC; PABC = 0
Let PA = Pb = PC = a < 1 / 2
P(A∪B∪C)=a+a+a-a²-a²-a²=9/16
It is found that a = 1 / 4,3 / 4 (greater than 1 / 2)
So a = 1 / 4