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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.

 Aug 8, 2023
 #1
avatar+128398 
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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 

 

cool cool cool

 Aug 8, 2023
 #2
avatar+52 
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Thank you so much! I forgot to mention but I believe the answer should be in the form of a common fraction. However, I'm not sure how I should convert it. (It's a program where one inputs the answer and doesn't take sin/cos etc. in the expression)  

choutowne  Aug 8, 2023
 #4
avatar+52 
0

Nevermind, thanks, I got the answer \(\boxed{\frac{16}{7}}\)! Thanks for your help!

choutowne  Aug 10, 2023

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