Knockout JEE Main April 2021 (One Month)
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Let the height of the tower be h
so
the distance of the point A from the foot of tower be b
so so tan 45 =
⇒ b = x {tan 45° = 1}
for ΔABC
so tan 60° =
⇒ √3x = x + 5
⇒ 0.732x = 5
⇒ x ≈ 6.83 m