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xmistermayhem

  • one year ago

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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  1. anonymous
    • one year ago
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    start by drawing a picture. what do you think this scenario looks like?

  2. sohailiftikhar
    • one year ago
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    |dw:1439384614923:dw|

  3. sohailiftikhar
    • one year ago
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    \[\tan \theta = x/60\]

  4. xmistermayhem
    • one year ago
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    What about the 6 feet, though?

  5. sohailiftikhar
    • one year ago
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    nothing we will add after knowing the value of x

  6. xmistermayhem
    • one year ago
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    Oh okay Uhhh... How do you find that? What does θ mean?

  7. sohailiftikhar
    • one year ago
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    |dw:1439384809225:dw|

  8. sohailiftikhar
    • one year ago
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    \[\theta=45\]

  9. xmistermayhem
    • one year ago
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    So whatever x is + 6 = the height of the building?

  10. sohailiftikhar
    • one year ago
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    yes! you got it buddy

  11. xmistermayhem
    • one year ago
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    so tan 45 degrees = x/60 ?

  12. sohailiftikhar
    • one year ago
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    it's 66 feet

  13. sohailiftikhar
    • one year ago
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    yes

  14. xmistermayhem
    • one year ago
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    Ohhh okay Thank you!!! I think I got it now!! : )

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