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

Mathematics
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welcome!!!
In triangle DEF, what is the length of DF?
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here is the question.....i am not really sure how to do it??

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

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