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jazy

  • one year ago

I'll put my question in the comments.

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  1. jazy
    • one year ago
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    How would I find AB on here: |dw:1365267611167:dw|

  2. jazy
    • one year ago
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    If they give me these length measures. |dw:1365267723624:dw|

  3. jazy
    • one year ago
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    They're both right triangles. I have the lengths of the sides, but I want someone to explain how I could find AB, without giving me the answer.

  4. AravindG
    • one year ago
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    sorry but the side lengths are not readable :/

  5. AravindG
    • one year ago
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    or were they arrows ? :)

  6. Rav
    • one year ago
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    This is a matter of using Pythagoras' theorem. a^2 + b^2 = c^2 so name your sides, and then solve for all the things you know. Even if you don't know the side lengths, you can get a formula for the length of AB.

  7. jazy
    • one year ago
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    Yeah lol and bad ones too. Do you need the length measures to help me?

  8. AravindG
    • one year ago
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    I would recommend you to use similar triangles here

  9. jazy
    • one year ago
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    I tried the Pythagoras theorem but neither one of my triangles have 2 side lengths given. :/

  10. AravindG
    • one year ago
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    |dw:1365268463728:dw|

  11. Rav
    • one year ago
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    If there are no side lengths given then you need to find a formula. Instead of substituting the side length for the variable, just keep the variable in there.

  12. jazy
    • one year ago
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    Sorry, my brain is completely blank today when it comes to geometry. I'm not sure what you mean.

  13. AravindG
    • one year ago
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    hmm...

  14. AravindG
    • one year ago
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    I need to clarify is CBA right angled?

  15. jazy
    • one year ago
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    Yes: |dw:1365268759222:dw|

  16. AravindG
    • one year ago
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    Are u sure this is only the info given ?

  17. jazy
    • one year ago
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    These aren't the ones given to me but maybe they'll work: |dw:1365268927327:dw|

  18. AravindG
    • one year ago
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    why not try this ? |dw:1365268967361:dw|

  19. AravindG
    • one year ago
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    AB is the only unknown

  20. jazy
    • one year ago
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    Ok, I'll try it out on paper first. One sec..

  21. AravindG
    • one year ago
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    ok :)

  22. AravindG
    • one year ago
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    waiting....

  23. jazy
    • one year ago
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    I don't know what wrong with me today :/ So I get: \[\tan(A) = \frac{ 104 }{ x} = \frac{ 186 }{ x + 98 }\]But how could I solve this with two variables?? o.o

  24. Rav
    • one year ago
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    You only have one variable. A is irrelevant, all you want is x.

  25. jazy
    • one year ago
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    Oh, ok. So would I cross multiply now to solve for x?

  26. Rav
    • one year ago
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    Yes. Please give your medals to aravind :)

  27. AravindG
    • one year ago
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    :)

  28. jazy
    • one year ago
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    Ok and thanks(: \[\frac{ 104 }{ x } = \frac{ 186 }{ x + 98 } \to 104(x + 98) = 186x \to 104x + 192 = 186x \to 192 = 82x \to 2.34\]

  29. AravindG
    • one year ago
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    |dw:1365270145268:dw|

  30. AravindG
    • one year ago
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    what happened @jazy ?you seem terrible on math today :/

  31. jazy
    • one year ago
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    @AravindG I wish I knew. -.- Usually I'm pretty quick to catch on, but I guess today just isn't my day for math. I'll switch subjects for a while to refresh. Thanks for the patience and help, I really do appreciate it. c:

  32. AravindG
    • one year ago
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    yw :)

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