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Thanks in advance.

Guest May 7, 2018
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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°

 

 

 

cool cool cool

CPhill  May 7, 2018

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