regular tetrahedron with base ABCD has a point in its interior we call P such that,
AP=BP=sqrt(11)
CP=DP=sqrt(17)
AB=BC=CD=AD=x
what is the value of x?
I have tried:
if we let theta be <APB, phi be <APD, and omega be <PCD
we have using law of cosines
(i)x^2 =2*11-2(11)cos(theta)
(ii)x^2 =11+17-2sqrt(11)sqrt(17)cos(phi)
(iii)x^2 =2*17-2(17)cos(omega)
(iv)17 =17+x^2-2sqrt(17)xcos(pi/2-omega/2)
iv-->v)17 =17+x^2-2sqrt(17)xsin(omega/2)
Where to go from here? Four equations, four unknowns...