1.

A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in fron of the tower C and spots a very distant objct O in line with AC. He then walks pependicular to AC up to B, a distance of 100 m and look at O and C again. Since O is very distant the direction BO is practivally the same as AO, but he finds the line of slight by of C shifted fro the original line of slight by an angle theta=30^(@) (theta is known as parallax) estimate the distance of the tower C from his original position A.

Answer»

SOLUTION :
Parallax ANGLE `theta=40^(@)`
`"tan" theta=(AB)/(AC)`
`:.AC=(AB)/(tan theta)`
`=(100)/(tan 40^(@))`
`=(100)/(0.8391)`
`=119.175`
`:.AC~~119 m`


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