+1622 If you know your basic trig laws, you will notice that LAW OF COSINES is perfect for this problem!
What is law of cosines? In a triangle ABC, with side lengths a, b, c (each opposite to the point), the formula is that:
c^2 = a^2 + b^2 - 2ab*cosC ... where angle C is included between a and b, and c is opposite point C. Without loss of generality, you can "mix and match" a, b, c.
We know DA = 3, and since triangle DEF is equilateral, then summing DA + AF = DF = DE = 5.
Using law of cosines:
\(AE^2 = AD^2 + DE^2 - 2*AD*DE*\cos{D} = 9 + 25 - 2*3*5*\cos{D}\)
D = 60 degrees, so cosD = cos(60 degrees) = adjacent/hypotenuse = 1/2
\(AE^2 = 9+25-2*3*5*{1\over{2}}=9+25-15=19\)
Square rooting both sides, you have AE = \(\sqrt{19}\)
+1622 If you know your basic trig laws, you will notice that LAW OF COSINES is perfect for this problem!
What is law of cosines? In a triangle ABC, with side lengths a, b, c (each opposite to the point), the formula is that:
c^2 = a^2 + b^2 - 2ab*cosC ... where angle C is included between a and b, and c is opposite point C. Without loss of generality, you can "mix and match" a, b, c.
We know DA = 3, and since triangle DEF is equilateral, then summing DA + AF = DF = DE = 5.
Using law of cosines:
\(AE^2 = AD^2 + DE^2 - 2*AD*DE*\cos{D} = 9 + 25 - 2*3*5*\cos{D}\)
D = 60 degrees, so cosD = cos(60 degrees) = adjacent/hypotenuse = 1/2
\(AE^2 = 9+25-2*3*5*{1\over{2}}=9+25-15=19\)
Square rooting both sides, you have AE = \(\sqrt{19}\)
