A flagstaff is placed on top of a building. The flagstaff and building subtend equal angles at a point on level ground which is 200 m away from the foot of the building. If the height of the flagstaff is 50 m and the height of the building is h, which of the following is true?

Let AD be the flagstaff and CD be the building.
Assume that the flagstaff and building subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let ∠ABD = $$\theta $$, ∠DBC = $$\theta $$   (∵ flagstaff and building subtend equal angles at a point on level ground).
Then, ∠ABC = 2$$\theta $$
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta BCD, \cr & \tan \theta = \frac{{DC}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{200}}\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta BCA, \cr & \tan 2\theta = \frac{{AC}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AD + DC}}{{200}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + h}}{{200}} \cr} $$
$$ \Rightarrow \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{50 + h}}{{200}}$$      $$\left( {\because \tan \left( {2\theta } \right) = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$$
$$\frac{{2\left( {\frac{h}{{200}}} \right)}}{{1 - \frac{{{h^2}}}{{{{200}^2}}}}} = \frac{{50 + h}}{{200}}$$      (∵ substituted value of tan $$\theta $$ from eq:1)
$$\eqalign{ & \Rightarrow 2h = \left( {1 - \frac{{{h^2}}}{{{{200}^2}}}} \right)\,\left( {50 + h} \right) \cr & \Rightarrow 2h = 50 + h - \frac{{50{h^2}}}{{{{200}^2}}} - \frac{{{h^3}}}{{{{200}^2}}} \cr} $$
$$ \Rightarrow 2\left( {{{200}^2}} \right)h = 50{\left( {200} \right)^2} + $$     $$h{\left( {200} \right)^2} - $$   $$50{h^2} - {h^3}$$
(∵ multiplied LHS and RHS by $${{{200}^2}}$$ )

h³ + 50h² + (200)²h - (200)²50 = 0