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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 $AE$.

 

 Sep 10, 2023
 #1
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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

 

cool cool cool

 Sep 10, 2023

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