+964 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 $BE$.
+1908 Let's graph (yay!)
What's interesting about this is this is on a coordinate plane, so we can make quite a few deductions from it.
First off, we have that
\(CD = BD = 10 \\ ED = ED\)
Also though, note that ED and BC are just the x and y axis, thus they are perpendicular lines.
Which means that we have \(\angle CDE = \angle BDE\)
By SAS Congruence, we know that triangle CDE is congruent to triangle BDE.
Thus, we can conclude that corresponding sides \(CE = BE\)
Finally, we just simply right that
\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)
Thus, 10.35 SHOULD be correct, not sure, trig is never my thing :(
Thanks! :)
