The vertices of triangle $ABC$ lie on the sides of equilateral triangle $DEF$, as shown. If $CD = 5$, $CE = BD = 3$, and $\angle ACB = 90^\circ$, then find $AE$.
+129771 Law of Cosines
BC^2 = BD^2 + DC^2 - 2(BD * DC)cos (60°)
BC^2 = 3^2 + 5^2 - 2 (3*5)(1/2)
BC^2 = 34 - 15
BC^2 = 19
BC = sqrt (19)
BD^2 = BC^2 + DC^2 - 2(BC * DC)cos ( BCD)
3^2 = 19 + 5^2 - 2 ( sqrt (19)*5) cos (BCD)
35 / ( 10sqrt (19) = cos BCD
7 / (2sqrt 19) = 3.5 / sqrt (19) = cos BCD = sin ACE
arcsin (3.5 / sqrt 19) ≈ 53.41°
Angle CAE = 120 - 53.41 ≈ 66.59°
Law of Sines
CE /sin (CAE) = AC / sin (AEC)
CE /sin 66.59 = AC /sin 60
3 /sin 66.59 = AC/sin 60
AC = 3 sin 60 / sin 66.59 ≈ 2.83
AE / sin (ACE) = AC / sin (AEC)
AE / sin (53.41) = 2.83 / sin (60)
AE = 2.83 sin (53.41) / sin (60) ≈ 2.62
