A poster is on top of a building. A person is standing on the ground at a distance of 50 m from the building. The angles of elevation to the top of the poster and bottom of the poster are 45° and 30°, respectively. What is 200% of the height (in cm) of the poster?

Solution(By Examveda Team)

$$\eqalign{ & \tan {30^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{BD}}{{50}} \cr & \Rightarrow BD = \frac{{50}}{{\sqrt 3 }} \cr & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \cr & \Rightarrow AB = 50 \cr & \therefore AD = AB - BD \cr & = 50 - \frac{{50}}{{\sqrt 3 }} \cr & = \frac{{50\left( {\sqrt 3 - 1} \right)\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & = \frac{{50}}{3}\left( {3 - \sqrt 3 } \right){\text{m}} \cr & 200\% {\text{ of }}AD = \frac{{50}}{3}\left( {3 - \sqrt 3 } \right) \times \frac{{200}}{{100}} \cr & = \frac{{100}}{3}\left( {3 - \sqrt 3 } \right){\text{m}} \cr} $$