# A tower is 'b ' metre high having a flagstaff at its top.The tower and flagstaff subtend equal angles at a point of distance 'a'm from foot of tower.Prove:{b(a^2+b^2)}/(a^2-b^2)=LENGTH OF FLAGSTAFF

embizze | High School Teacher | (Level 2) Educator Emeritus

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Let b be the height of the tower, a the distance to the point where the angles subtended are equal, and l the height of the flagpole.

Let `theta` be the angle subtended by the tower. Then the angle subtended by the flagpole is `theta` and the angle subtended by the tower and the pole is `2theta` .

Now `tan theta=b/a` using right triangles. Also `tan2theta=(b+l)/a` .

Using `tan2theta=(2tantheta)/(1-tan^2theta)` we get:

`(b+l)/a=tan2theta=(2b/a)/(1-(b/a)^2)`  ** multiply by `a^2/a^2` to get

`=(2ab)/(a^2-b^2)`   So `(b+l)/a=(2ab)/(a^2-b^2)`

`==>b+l=(2a^2b)/(a^2-b^2)`

`==>l=(2a^2b)/(a^2-b^2)-b`

`==>l=(2a^2b-a^2b+b^3)/(a^2-b^2)`  Get common denominator and add

`==>l=(a^2b+b^3)/(a^2-b^2)`

`==>l=(b(a^2+b^2))/(a^2-b^2)` as required.