+58 The vertices of triangle ABC lie on the sides of equilateral triangle DEF, as shown below in the image. If CD = 5, CE = BD = 2, and angle ACB = 90 degrees, then find AE.
+129881 Using the Law of Cosines
BC^2 = DB^2 + DC^2 - 2(DB*DC) cos (60 degrees)
BC^2 = 2^2 + 5^2 - 2 (2*5)(1/2)
BC^2 = 29 - 10
BC = sqrt (19)
Using the Law of Sines
sin BCD / BD = sin BDC / BC
sin BCD / 2 = sqrt (3) / [ 2 sqrt (19) ]
sin BCD = sqrt (3/19)
BCD = arcsin (3/19) ≈ 23.4°
Angle ACE = 90 - 23.4 ≈ 66.6°
Angle CAE = 180 -60 - 66.6 ≈ 53.4°
Using the Law of Sines again
CE / sin CAE = AE /sin ACE
2 / sin (53.4) = AE / sin (66.6)
AE = 2sin (66.6) / sin (53.4) ≈ 2.286
