A vertical tower PQ subtends the same angle of 30 degree at each of the two places A and B, 60 m apart on the ground. If AB subtends an angle of 120 degree at P, the foot of the tower, find the height of the tower.

20 m

How

In the 3D figure; 
tan30=PQ/AP​
1/(3)^(1/2)  = PQ/AB 
AP= PQ* √3.
Similarly;
PB=h√3.     Where PQ= h
So;.  AP=  PB
So; TRIANGLE APB is isosceles
ang. PAB= ang. PBA= 30

Now using law of sines;
AP/sin30 = AB/sin120
AP/(1/2) = 60/√3/2
AP = h√3 so;
h√3*√3= 60
3h=60
And hence,
h=20

CaN draw the figure Please?

Can u Draw the figure plz?

hi
 

its 20m
 

In the 3D figure;?
tan30=PQ/AP?
1/(3)^(1/2) ?= PQ/AB?
AP= PQ* ?3.
Similarly;
PB=h?3. ? ? Where PQ= h
So;. ?AP= ?PB
So; TRIANGLE APB is isosceles
ang. PAB= ang. PBA= 30

Now using law of sines;
AP/sin30 = AB/sin120
AP/(1/2) = 60/?3/2
AP = h?3 so;
h?3*?3= 60
3h=60
And hence,
h=20

AP=pB=hcot30°=√3h 1/2=cos120°=3h*+3h*-60*/2×3h* +3h*=3h*+3h*-3600 9h*=3600 h=20 * Means square