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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 $BE$.

 May 23, 2024
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So we know that

 

\(CD = BD = 10 \\ ED = ED\)

 

Through this, we can find that triangle CDE is congruent to triangle BDE.

This means BE = CE!

 

 

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

 

Thanks! :)

 May 23, 2024

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