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?

Answer: Option D

Solution

Answer: Option D

Solution:

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