An engineer is surveying a building with a device that is 6 feet above the ground. He is 60 feet from the building and is looking with an angle of elevation of 45°. How tall is the building? ... I'm really bad at trig. x....x

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An engineer is surveying a building with a device that is 6 feet above the ground. He is 60 feet from the building and is looking with an angle of elevation of 45°. How tall is the building? ... I'm really bad at trig. x....x

Mathematics
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start by drawing a picture. what do you think this scenario looks like?
|dw:1439384614923:dw|
\[\tan \theta = x/60\]

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What about the 6 feet, though?
nothing we will add after knowing the value of x
Oh okay Uhhh... How do you find that? What does θ mean?
|dw:1439384809225:dw|
\[\theta=45\]
So whatever x is + 6 = the height of the building?
yes! you got it buddy
so tan 45 degrees = x/60 ?
it's 66 feet
yes
Ohhh okay Thank you!!! I think I got it now!! : )

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