anonymous
  • anonymous
I have a Geometry question if anybody can help!! :) thnx!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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shubhamsrg
  • shubhamsrg
welcome!!!
anonymous
  • anonymous
In triangle DEF, what is the length of DF?
1 Attachment
anonymous
  • anonymous
here is the question.....i am not really sure how to do it??

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terenzreignz
  • terenzreignz
Well, for starters, DF is the hypotenuse of this right triangle, right? :)
anonymous
  • anonymous
yes!!
terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
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??
terenzreignz
  • terenzreignz
Hang on :) Is this side... |dw:1362666135050:dw| Opposite to the 60 degree angle? Is it adjacent to it?
terenzreignz
  • terenzreignz
Oh wait, it seems like you already got the answer down...
anonymous
  • anonymous
wait what??
terenzreignz
  • terenzreignz
Okay, forget what I said. Do you use sines and cosines already? :D
anonymous
  • anonymous
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??
terenzreignz
  • terenzreignz
Okay, you want to use sines and cosines, or just the properties of the so-called 30-60-90 triangles?
anonymous
  • anonymous
YES!!!
terenzreignz
  • terenzreignz
Yes, what? I gave you a choice :P
anonymous
  • anonymous
oo sorry I only looked at the last one which is 30-60-90 traingles that is what i wanted
terenzreignz
  • terenzreignz
LOL OK :) so... suppose we have a 30-60-90 triangle |dw:1362666525444:dw|
anonymous
  • anonymous
okay :)
terenzreignz
  • terenzreignz
Now, if one side measures...|dw:1362666670884:dw| This is the side opposite the 30-degree angle, right?
anonymous
  • anonymous
yes!
terenzreignz
  • terenzreignz
Well, in this case..., the angle opposite the 60-degree angle would measure... |dw:1362666746027:dw| And the hypotenuse would measure
terenzreignz
  • terenzreignz
|dw:1362666784331:dw|
anonymous
  • anonymous
umm....well I guess 45??
terenzreignz
  • terenzreignz
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.
terenzreignz
  • terenzreignz
*hypotenuse
anonymous
  • anonymous
wait this is were I get confused is adding them up!!
terenzreignz
  • terenzreignz
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
anonymous
  • anonymous
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!!
terenzreignz
  • terenzreignz
No problem :)
terenzreignz
  • terenzreignz
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 :)
anonymous
  • anonymous
yeah!! I will write down those steps you gave me and use them in the future!! thank you :)

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