In the question A, B and O are in a horizontal plane and P is vertically above O, where OP=h m. Find the value of h.A is due west of O, B is due...

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