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| Question 1193619:  Given:	right △DEF with
 m∠E = 2 · m∠F and EF = 14radical 3
 Find:	DE and DF
 △D E F is given. ∠D is a right angle.
 simplest radical form	DE	=
 
 approximation	DE	=
 simplest radical form	DF	=
 
 approximation	DF	=
 
 
 Found 2 solutions by  Boreal, MathTherapy:
 Answer by Boreal(15235)
      (Show Source): 
You can put this solution on YOUR website! DE is half of the hypotenuse or 7 sqrt (3). It is because it is a 30-60-90 right triangle, which is the only kind where one of the angles is half the other. -
 DE is 7*1.73= 12.21
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 DF is the short leg * sqrt (3)=21
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 it is a 7 sqrt(3)//21//14 sqrt(3) right triangle or squaring each, 147+441=196*3=588
Answer by MathTherapy(10556)
      (Show Source): 
You can put this solution on YOUR website! Given:	right △DEF with m∠E = 2 · m∠F and EF = 14radical 3
 Find:	DE and DF
 △D E F is given. ∠D is a right angle.
 simplest radical form	DE	=
 approximation	DE	=
 simplest radical form	DF	=
 approximation	DF	=
 
 AGAIN, 2 scenarios exist. 1) ∡s E, D, and F can be 90o, 45o, and 45o, respectively.
 2) ∡s E, D, and F can ALSO be 60o, 90o, and 30o, respectively.
 This triangle can therefore be a 45-45-90 special triangle, or one of the 30-60-90 variety, since DE and DF can EACH have 2 different lengths.
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