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 m) of the poster?

$\frac{100}{3}(3 - \sqrt{3})$

⇒ In triangle BCD, tan \({30}^\circ\) = \(\frac{BC}{CD}\)

We know that tan \({30}^\circ\) = \(\frac{1}{√3}\) and CD = 50m,

⇒ \(\frac{BC}{50}\) = \(\frac{1}{√3}\)

⇒ BC = \(\frac{50}{√3}\) or \(\frac{50√3}{3}\)

Similarly in ACD

tan \({45}^\circ\) = \(\frac{AC}{CD}\)

⇒ AC = AB + BC, tan \({45}^\circ\) = 1 and CD = 50m

⇒ AC = CD or AB + BC = CD

Substituting the values we get,

⇒ AB + \(\frac{50√3}{3}\) = 50,

⇒ AB= 50 - \(\frac{50√3}{3}\)

⇒ AB = 50 - \(\frac{50√3}{3}\)

⇒ AB = \(\frac{50}{3}\)(3 - \(\sqrt {3 }\))

Therefore, 200% of the Poster height = \(\frac{100}{3}\)(3 - \(\sqrt {3 }\))