Respuesta :
Refer to the attached image.
Let the height of the Empire State Building = BC = 'x'
John is standing at distance from Empire State Building = AB = 1 mile
The angle of elevation = 11°
In triangle ABC,
Consider [tex] \tan 11^{\circ}= \frac{Perpendicular}{Base} [/tex]
[tex] \tan 11^{\circ}= \frac{BC}{AB} [/tex]
[tex] \tan 11^{\circ}= \frac{x}{1} [/tex]
[tex] 0.194= \frac{x}{1} [/tex]
x = 0.194 mile
Since 1 mile =5280 feet
So, x = [tex] 0.194 \times 5280 [/tex]
= 1024.32 feet.
So, the height of the Empire State Building is 1024.32 feet.
![Ver imagen pinquancaro](https://us-static.z-dn.net/files/d71/f5d75d5daae561a7354f24cfd0af9c2c.png)