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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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