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mark this one also as brainliest please...... please please please
Solution :-
Let AB be the height of the tower and DE be the height of the observer.
Then,
In Δ ACD ,AC/DC
= tan 30°
⇒ x/√30
= tan 30°
= 1/√3 ⇒
x = 20 mAB
= 20 m + 1.7 m= 21.7 m
So, the height of the tower is 21.7 m