Hlw frnds....Plz answer it as fast as possible...
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Hi friend,
# 43.30m
➡Complete solution :-
Refer to this diagram ,
Let the the tower = AB
BC is original shadow
BD is final shadow
Given BD-BC = CD = 50m
Therefore , in ABC,
tan 60° = AB/BC
=> AB/root 3 = BC.....i)
➡Therefore, In ABD
tan 30° = AB/BD
Or, 1/root 3 = AB/(BC +CD)
Or , ABroot 3 = AB/root 3+50..(1)
Or, AB root 3 - AB root 3 = 50
Or, 3 AB-AB root 3 = 50
Or , AB = 50 root 3/2
= 25 root 3
= 25*1.732 = 43.30m
...i hope it helps you
_____________________________< Mark me as a brainliest Be brainly !! Thanks
# 43.30m
➡Complete solution :-
Refer to this diagram ,
Let the the tower = AB
BC is original shadow
BD is final shadow
Given BD-BC = CD = 50m
Therefore , in ABC,
tan 60° = AB/BC
=> AB/root 3 = BC.....i)
➡Therefore, In ABD
tan 30° = AB/BD
Or, 1/root 3 = AB/(BC +CD)
Or , ABroot 3 = AB/root 3+50..(1)
Or, AB root 3 - AB root 3 = 50
Or, 3 AB-AB root 3 = 50
Or , AB = 50 root 3/2
= 25 root 3
= 25*1.732 = 43.30m
...i hope it helps you
_____________________________< Mark me as a brainliest Be brainly !! Thanks
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SwetaDey:
thnk you very mch
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