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In triangle $ABC$, let the perpendicular bisector of $BC$ intersect $BC$ and $AC$ at $D$ and $E$, respectively.  If $BC = 20$ and $\angle C = 15^\circ$, then find the length of $BE$.

 Oct 12, 2023
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Triangle ABC is  isosceles with   AB  = AC

Angle ADB = 90

Angle ABD = 15

Angle BAD = 75 

 

AB  =  10 / sin 75°  =  10 / cos 15°

AE = EC  =  5 / cos 15°

 

BE^2  = BC^2  + EC^2  -  2 ( BC * EC) cos (15°)

 

BE^2 =  20^2  +  25 / [cos 15°]^2  - 2 ( 20 * 5/ cos (15°) * cos 15°

 

BE^2  = 400 + 25 / [ cos 15°]^2  -  200

 

BE^2  =  200 + 25 / [ cos 15°]^2    { [ cos 15°]^2  =   (2 + sqrt 3)/4

 

BE^2  = 200 + 25 * 4 / ( 2 + sqrt 3)

 

BE^2  = 200 + 100 / (2 + sqrt 3)

 

BE^2   =  100  ( 2  +  1/ (2 + sqrt 3) )

 

BE^2  = 100  ( 2  + 2 - sqrt 3 )

 

BE^2  =  100 ( 4 -sqrt 3)

 

BE  = 10 sqrt ( 4 -  sqrt 3)

 

 

cool cool cool 

 Oct 13, 2023

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