We have three points A, B and O in a plane and point P lies a distance h above O. A is due west of O, B is due South of O and AB= 60m. The angle of elevation of P from A is 25 degrees and the angle of elevation of P from B is 33 degrees.

We have three right triangles here: AOB, AOP and BOP.

As the angle of elevation of P is 25 degrees from A, we get:

tan 25 = h/AO

=> AO = h/tan 25

As the angle of elevation of P from B is 33 degrees, we get:

tan 33 = h/OB

=> OB = h/tan 33

Now AB^2 = AO^2 + OB^2

=> 60^2 = h^2/(tan 25)^2 + h^2/(tan 33)^2

=> h^2 = 60^2*(tan 25)^2*(tan 33)^2/[(tan 25)^2 + (tan 33)^2]

=> h^2 = 516.49

h = sqrt (516.49) = 22.72 m

**The height of the point P, h = 22.72 m**

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