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space4cadet101

  • one year ago

I have a Geometry question if anybody can help!! :) thnx!!!

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  1. shubhamsrg
    • one year ago
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    welcome!!!

  2. space4cadet101
    • one year ago
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    In triangle DEF, what is the length of DF?

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  3. space4cadet101
    • one year ago
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    here is the question.....i am not really sure how to do it??

  4. terenzreignz
    • one year ago
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    Well, for starters, DF is the hypotenuse of this right triangle, right? :)

  5. space4cadet101
    • one year ago
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    yes!!

  6. terenzreignz
    • one year ago
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    Yes... let's redraw this figure, but only including the relevant stuff... |dw:1362665968423:dw| Now, with respect to the sixty degree angle, what is the side measuring 45?

  7. space4cadet101
    • one year ago
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    I am not really sure I would think the other angle down below would be 30 degrees if there was one?? I would guess it would be 30 square root of three??

  8. terenzreignz
    • one year ago
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    Hang on :) Is this side... |dw:1362666135050:dw| Opposite to the 60 degree angle? Is it adjacent to it?

  9. terenzreignz
    • one year ago
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    Oh wait, it seems like you already got the answer down...

  10. space4cadet101
    • one year ago
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    wait what??

  11. terenzreignz
    • one year ago
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    Okay, forget what I said. Do you use sines and cosines already? :D

  12. space4cadet101
    • one year ago
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    yes I did but can you give me the math part of the equation like how to do it?? I mean I think I know how to do it but I need someone to just write out how they did the equation! if that makes any since??

  13. terenzreignz
    • one year ago
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    Okay, you want to use sines and cosines, or just the properties of the so-called 30-60-90 triangles?

  14. space4cadet101
    • one year ago
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    YES!!!

  15. terenzreignz
    • one year ago
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    Yes, what? I gave you a choice :P

  16. space4cadet101
    • one year ago
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    oo sorry I only looked at the last one which is 30-60-90 traingles that is what i wanted

  17. terenzreignz
    • one year ago
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    LOL OK :) so... suppose we have a 30-60-90 triangle |dw:1362666525444:dw|

  18. space4cadet101
    • one year ago
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    okay :)

  19. terenzreignz
    • one year ago
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    Now, if one side measures...|dw:1362666670884:dw| This is the side opposite the 30-degree angle, right?

  20. space4cadet101
    • one year ago
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    yes!

  21. terenzreignz
    • one year ago
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    Well, in this case..., the angle opposite the 60-degree angle would measure... |dw:1362666746027:dw| And the hypotenuse would measure

  22. terenzreignz
    • one year ago
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    |dw:1362666784331:dw|

  23. space4cadet101
    • one year ago
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    umm....well I guess 45??

  24. terenzreignz
    • one year ago
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    What? I didn't ask you anything yet... :P But yeah, you're given the side opposite the 60 degree angle, so that's 45. And this is equal to x sqrt(3) So... \[\large 45=x\sqrt3\] Now solve for x. Actually, solve for 2x, since that's the measure of the hypontenuse.

  25. terenzreignz
    • one year ago
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    *hypotenuse

  26. space4cadet101
    • one year ago
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    wait this is were I get confused is adding them up!!

  27. terenzreignz
    • one year ago
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    Okay, slowly... \[45=x\sqrt3\]Now multiply both sides by 2 \[90=2x\sqrt3\]Now divide both sides by sqrt(3) \[\large \frac{90}{\sqrt{3}}=2x\]And rationalise the denominator... \[\large \frac{90\sqrt{3}}{3}=2x\]\[\huge 30\sqrt3=2x\]And so you were right the first time :D

  28. space4cadet101
    • one year ago
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    okay wow that is really nice of you to write all of that out for me :) thank you so much!! I will give you a medal for all of your help!! thank you :) have a great day!!

  29. terenzreignz
    • one year ago
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    No problem :)

  30. terenzreignz
    • one year ago
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    But make sure you got how to do this... it's really a rewarding feeling, solving one type of problem for the first time on your own... real satisfying :)

  31. space4cadet101
    • one year ago
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    yeah!! I will write down those steps you gave me and use them in the future!! thank you :)

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