Question

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45. Find the height of the tower PQ and the distance PX. 

Answer 1

Case 1:
Angle = 60° height =PQ Base = PX
tan 60= PQ/PX
PQ=√3PX-------(1)

CASE 2:
Angle = 45° height =(PQ - 40)m Base = PX
tan 45 = (PQ - 40)/PX
PX=PQ-40
PX = √3PX - 40 (BY APPLYING equation 1)
PX = 40/0.732
PX= 54.64 m

In eq 1
PQ=√3 PX
PQ= 94.64m

Answer 2

in triangle YRQ,
   
           tan 45 =  QR/YR
              1=  z/YR
                YR = z

in triangle XPQ
                     TAN 60 = PQ/PX
                        root 3 = z+40/z
                        root 3 x z = z + 40
                      root 3 x z - z = 40 
                           z = 40 / ( root 3 -1)
PQ = 20 x ( root 3 - 1) + 40 = 94.64

PX = tan 60 = QP/XP 
     PX = 31.54 X Root 3 

hope it helps!