I doubt that GMAT will ever use angles like 38 and 29 .

The solution to the problem can be expressed in trigonometric terms though.
On a 3D coordinate:
If O is the origin and AB (hypotenuse ) = 50 , its other sides will be OB = 40 (smaller angle means that the point of sight is farther) and OA=30 (greater angle means that the point of sight is nearer )

Now we can use any of these sides to determine the height of the tower
could be tan(38) = H/30 or tan(29) =H/40 ,where H is the Height of the tower.

Btw whats the source?

marmar29 wrote:the angle of elevation of the top of a tower is 38 from a point A due to south of it
the angle of elevataion of the top of the tower from another point B ,due to east is 29. find tht ht of the tower if the distance AB is 50 m
pls show the stps
manar

thanks alooot for yr great help but i notice tht u put thte rule for tan 38 = H/ 30
BUT THE TAN = O/A
CAN U CLARIFY HOW U USE HYPOTENUS IN THE RULE ???

see the diagram

marmar29 wrote:thanks alooot for yr great help but i notice tht u put thte rule for tan 38 = H/ 30
BUT THE TAN = O/A
CAN U CLARIFY HOW U USE HYPOTENUS IN THE RULE ???

a great work !! thanks manar

marmar29 wrote:a great work !! thanks manar

but whn i tried to solve it tan 29 if gves another final other thn tan 38 !!
isnt it shud have the sme ht in any way !!

I think angles should have been different.GMAT questions will never contradict answers. Btw whats the source?

marmar29 wrote:

marmar29 wrote:a great work !! thanks :) manar

but whn i tried to solve it tan 29 if gves another final other thn tan 38 !!
isnt it shud have the sme ht in any way !!