At the top of their 50 foot tall sales building, there is a 10 feet tall sign. But he wants to replace it with a sign for which the ideal viewing distance is 60 feet from the building. For an observer, the angle between the lines of sight from the observer's eye to the top and the bottom of the sign is a measure of the observer's view of the sign, and the best view occurs when this angle is largest possible. Assuming that the observer's eyes are 5 ft off the ground (so that the observer is approximately 5'3"), you can determine that the ideal viewing distance from the building is approximately 49.7 feet if the sign were 10 feet tall. You are to determine the height of the new sign placed at the top of the building so that the best view occurs when the observer is 60 feet from the building. Let h be the height of the new sign, and let x be the distance that the observer stands from the building. So here is what I have done: Tan theta= 50/x Theta = arctan 50/x Tan (theta + beta) = 50 + h / x (theta + beta) = arctan 50 + h / x theta = arctan 50 + h / x – arctan 50 / x theta ‘ = 1/ 1 + (50 + h / x)^2 (-50 – h/ x^2) – 1/ 1 + (50/x)^2 (-50/x^2) = -50 – h / x^2 + (50 + h)^2 + 50 / x^2 + 50^2 = 0 50 / x^2 + 50^2 = 50 + h / x^2 + (50 + h)^2 I hope you can understand, I can't get a triangle on here to show you so I hope you can follow along. Once I get to the last line, i'm not sure what to do. I've tried plugging 60 in for x but i don't think I get a correct answer.