Asked by Meenakshi
The angle of elevation of the top of a tower as observed from a point on the ground is alpha and on moving a meters towards the tower, the angle of elevation is beta .prove that the height of the tower is a (tan alpha*tan beta)/(tan beta - tan alpha)
Answers
Answered by
Reiny
make a sketch, label the height as h
distance from last position to the tower as x
tan (alpha) = h/(x+a)
h = (x+a)tan alpha
in the same way
h = xtan beta
(x+a)tan alpha = xtan beta
x tan alpha - xtan beta = - a tan alpha
x(tan alpah - tan beta) = -a tan alpha
x = -a tan alpha/(tan alpha - tan beta)
x = a tan alpha /(tan beta - tan alpha)
h = xtan beta
= a tan alpha * tan beta/(tan beta - tan alpha)
as requested
distance from last position to the tower as x
tan (alpha) = h/(x+a)
h = (x+a)tan alpha
in the same way
h = xtan beta
(x+a)tan alpha = xtan beta
x tan alpha - xtan beta = - a tan alpha
x(tan alpah - tan beta) = -a tan alpha
x = -a tan alpha/(tan alpha - tan beta)
x = a tan alpha /(tan beta - tan alpha)
h = xtan beta
= a tan alpha * tan beta/(tan beta - tan alpha)
as requested
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