A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant , the direction BO is practically the same as AO, but he finds the line of sight of C shifted from the original line of sight by an angle theta = 40^(@) (theta is known as 'Parallax') estimate the distance of the tower C from his original position A.

A man wishes to estimate the distance of a nearby tower from him. He s

1.	A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant , the direction BO is practically the same as AO, but he finds the line of sight of C shifted from the original line of sight by an angle theta = 40^(@) (theta is known as 'Parallax') estimate the distance of the tower C from his original position A.
Answer» Solution :We have, PARALLAX angle `THETA = 40^(@)` From fig.2.3. AB=AC TAN`theta` AC=AB/tan`theta` = 100m/tan`40^(@) = 100m//0.8391 = 119`