sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
we have the information that From a point P on level ground,
the angle of elevation of the top of a tower is 300. If the tower is 100 m high,
now we have to find how far is P from the foot of the tower.
question no 6 , heights and distances , ICSE board, ML Aggarwal
Consider AB as the tower and P is at a distance of x m from B which is the foot of the tower.
Height of the tower = 100 m
Angle of elevation = 300
We know that
tan θ = AB/PB
Substituting the values
tan 300 = 100/x
So we get
1/√3 = 100/x
By cross multiplication
x = 100√3
x = 100 (1.732) = 173.2 m
Hence, the distance of P from the foot of the tower is 173.2 m.