Question

hey guys Pls draw the apt diagram and solve on paper

Answer 1

(Let height of tower b h metres.)

tan30=PQ/AP​
1/(3)^(1/2)  = PQ/AB 
AP= PQ* √3.

Similarly,
PB=h√3.    (PQ= h)
 AP=  PB
TRIANGLE APB is isosceles
ang. PAB= ang. PBA= 30

using law of sines
AP/sin30 = AB/sin30
AP/(1/2) = 60/(1/2)
AP = h√3
h√3= 60
h=60/√3
=60√3/3
h=20√3m

Answer 2

sorry friend my attachment clip is not opening so I am just reading the solution I am really sorry for the diagram.
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Hi friend,
•••••••••••
Method of solution :-
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Tan 30=PQ/AP
1/3½=PQ/AB
AP =PQ×√3
Then,
PB =h√3 (°•°Given A . T. Q . )
AP =PB
∆APB is the angle of isosceles which is
/_PAB=/_PBA=30°
The property of sines
AP /sin 30=AB/sin30
AP /(1/2)=60/(1/2)
AP=h√3
h√3=60 [AP=60]
h =60/√3
h=60√3/3
h=20√3

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