A. AB is the hypotenuse of a 30-60-90 right triangle....it will be twice as long as the side opposite the 30° angle = CB
So....AB = 2 CB = 2 (271) = 542 ft
B. We need to find AC...it will √3 times as long as BC = 271√3 ft
So...using the Law of Cosines....we have
AD^2 = AC^2 + DC^2 - 2 (AC) (DC) cos ( 74°)
AD = √ [ (271√3)^2 + (772)^2 - 2 (271√3)(772)cos (74°) ] = 785.2 ft
C. We can find angle E as follows :
DC^2 = DE^2 + CE^2 - 2(DE)(CE)cos (E)
772^2 = 543^2 + 561^2 - 2(543)(561) cos E
[ 772^2 - 543^2 - 561^2 ] / [ -2(543)(561) ] = cos E
Take the inverse cosne to find E
arccos ( [ 772^2 - 543^2 - 561^2 ] / [ -2(543)(561) ] ) = E = 88.7°