# 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 South of O and AB= 60m. The angle of elevation of P...

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 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.

If you want a reference to the diagram then please let me know as i will mail it to u as i don't know how to post a diagram on e-notes.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

By the details given AOB is a right angled triangle ( in a plane perpendicular to the plane OAB) with angle AOB = 90 deg.

AOP is a right angled triangle with AOP = 90 deg. Given OAP = 25 deg. So angle OPA = 90 - 25 = 65deg.  So OA = OP*tan65 = h tan 65.

Similarly from the right angled triangle OPB, OB = h tan (90-33) = h tan 57.

Now from the right angle triangle AOB, OA^2 + OB^2 = AB^2. Or

(h*tan 65)^2 + (h tan 57)^2 = 60^2 , as AB = 60 meter by data.

=> h^2 (tan^2 65+tan^2 57) = 60^2

=> h ^2 = 60^2/(tan^2 65 +tan^2 57)

h = 60/sqrt(tan^2 65 +tan^2 57)

h = 22.73 m.