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 $BE$.

jaekg Sep 18, 2024

#1**+1 **

Let's graph the question.

From the graph and the question, we know that

\(CD = BD = 10 \\ ED = ED \\ \angle CDE = \angle BDE = 90\)

So, by SAS congruence, we know that triangle CDE is congruent to triangle BDE

Thus, we can complete the problem in a series of steps using this congruence.

First, we know that

\(sin CED / CD = sin CDE / CE\)

Plugging in some numbers, we find that

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

Thus, 10.35 is approximately our answer.

Thanks! :)

*1800 points

NotThatSmart Sep 19, 2024