An engineer is surveying a building with a device that is 5 feet above the ground. He is 50 feet from the building and is lookin
g with an angle of elevation of 50°. How tall is the building?
1 answer:
1. Check the sketch of the problem.
2. In right triangle DCE,
![tan50= \frac{EC}{DC}= \frac{x}{50}](https://tex.z-dn.net/?f=tan50%3D%20%5Cfrac%7BEC%7D%7BDC%7D%3D%20%5Cfrac%7Bx%7D%7B50%7D%20%20)
tan 50 can be found to be 1.19 using a scientific calculator or other software.
![1.19=\frac{x}{50}](https://tex.z-dn.net/?f=1.19%3D%5Cfrac%7Bx%7D%7B50%7D)
x=1.19*50=59.5 (feet)
3. The height of the building is 59.5+5=64.5 (feet)
Answer: 64.5 (feet)
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