heights and distances

Asked by prgayathri | 26th Jan, 2009, 02:32: PM

Expert Answer:

Suppose NM is  the window.

EC is the opposite house.(which is taller than the window of course)

Let D be a point on EC such that angle END=A(angle of elevation)

and

 angle angle DNC= angle NCM = B

Also, let  ED=h, DC=x and DN=CM=y

Then,

 In triangle AED,

tan A =h/y...(1)

and

In triangle ACM,

 tanB= x/y...(2)

So equating the values of y frm both equations, we get,

h cot A = x cot B

So, we get,

h= xcotB/cotA

But the height of the building

=EC=ED+DC

=h+x

=(xcotB/cotA)+x

=x(tanA cot B+1)

Hence proved

 

Answered by  | 26th Jan, 2009, 07:08: PM

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