## space4cadet101 Group Title I have a Geometry question if anybody can help!! :) thnx!!! one year ago one year ago

1. shubhamsrg

welcome!!!

In triangle DEF, what is the length of DF?

here is the question.....i am not really sure how to do it??

4. terenzreignz

Well, for starters, DF is the hypotenuse of this right triangle, right? :)

yes!!

6. 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?

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

Hang on :) Is this side... |dw:1362666135050:dw| Opposite to the 60 degree angle? Is it adjacent to it?

9. terenzreignz

wait what??

11. terenzreignz

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??

13. terenzreignz

Okay, you want to use sines and cosines, or just the properties of the so-called 30-60-90 triangles?

YES!!!

15. terenzreignz

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

17. terenzreignz

LOL OK :) so... suppose we have a 30-60-90 triangle |dw:1362666525444:dw|

okay :)

19. terenzreignz

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

yes!

21. terenzreignz

Well, in this case..., the angle opposite the 60-degree angle would measure... |dw:1362666746027:dw| And the hypotenuse would measure

22. terenzreignz

|dw:1362666784331:dw|

umm....well I guess 45??

24. 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.

25. terenzreignz

*hypotenuse

wait this is were I get confused is adding them up!!

27. 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

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

No problem :)

30. 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 :)