SOLUTION: 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

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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.
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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(10552)   (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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