+1839 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 $BE$.
+129895 A E
B D C 15
10 10
EC / sin 90 = DC / sin75
EC = 10/sin75
BD = DC
ED = ED
So....by LL, right triangle BED congruent to right triangle CED
So BE = 10/sin75 ≈ 10.35
+16 I just want to add on that we can represent the answer \(10\sin(75) = 10\tan(15) = 10\tan(\frac{30}{2}).\)
Now we can use the trigonometric half-angle identity \(\tan(\frac{\alpha}{2}) = \frac{\sin(\alpha)}{\cos(\alpha)+1}\).
We can input our normal values: \(10\tan(\frac{30}{2}) = 10(\tan(\frac{30}{2})) = 10(\frac{\sin(30)}{\cos(30)+1}) = 10(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}+1) = 10(\frac{1}{\sqrt{3}+2}) = 10(2-\sqrt{3})= 20 - 10\sqrt{3}.\)
Therefore \(BE = \sqrt{(20 - 10\sqrt{3})^2+10^2} = \sqrt{400 + 300 - 400\sqrt{3} + 100} = \sqrt{800 - 400\sqrt{3}} = \sqrt{400(2-\sqrt{3})} = 20\sqrt{2-\sqrt{3}}\)
