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anonymous

  • 5 years ago

I have a question about trigonometry. I will write out the question. Once I post the question, I will then post an "answer" but it'd just be a drawing of how the triangle looks in the problem. Solve for x in the following right triangle, if a=53 and b=5.

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  1. anonymous
    • 5 years ago
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    |dw:1327031826563:dw|

  2. anonymous
    • 5 years ago
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    hopefully that is readable. I tried drawing that with my laptop.

  3. UnkleRhaukus
    • 5 years ago
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    well \[√a^2=x^2+(x+b)^2\]

  4. UnkleRhaukus
    • 5 years ago
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    \[a=x^2+x^2+2bx+b^2\] \[a=2x^2+2bx+b^2\]

  5. Mertsj
    • 5 years ago
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    |dw:1327032036209:dw|

  6. Mertsj
    • 5 years ago
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    |dw:1327032236614:dw|

  7. UnkleRhaukus
    • 5 years ago
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    triangles have positive side lengths

  8. Mertsj
    • 5 years ago
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    reject -7 since length of line segment can't be negative

  9. Mertsj
    • 5 years ago
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    so the result is x = 2

  10. UnkleRhaukus
    • 5 years ago
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    that is a 2,7,√53 triangle

  11. anonymous
    • 5 years ago
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    So you're allowed to get the answer to equal 0, correct? (where you subtracted the 53 from the right side)

  12. UnkleRhaukus
    • 5 years ago
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    which we know is correct because \[2^2+7^2=√53^2\]

  13. anonymous
    • 5 years ago
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    So you're allowed to get the answer to equal 0, correct? (where you subtracted the 53 from the right side)

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