a) h³ – 50h² + (200)²h + (200)²50 = 0
b) h³ – 50h² – (200)²h + (200)²50 = 0
c) h³ + 50h² + (200)²h – (200)²50 = 0
d) None of these

Answer: c
Explanation:

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)
$$\eqalign{ & 2h = \left( {1 – \frac{{{h^2}}}{{{{200}^2}}}} \right)\,\left( {50 + h} \right) \cr & 2h = 50 + h – \frac{{50{h^2}}}{{{{200}^2}}} – \frac{{{h^3}}}{{{{200}^2}}} \cr} $$
$$ 2\left( {{{200}^2}} \right)h = 50{\left( {200} \right)^2} + $$     $$h{\left( {200} \right)^2} – $$   $$50{h^2} – {h^3}$$
( multiplied LHS and RHS by $${{{200}^2}}$$ )

h³ + 50h² + (200)²h – (200)²50 = 0