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# Geometry

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+1646

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 $BE$.

Jul 21, 2024

#1
+129806
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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

Jul 21, 2024
#2
+16
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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 $$BE = \sqrt{(20 - 10\sqrt{3})^2+10^2} = \sqrt{400 + 300 - 400\sqrt{3} + 100} = \sqrt{800 - 400\sqrt{3}} = \sqrt{400(2-\sqrt{3})} = 20\sqrt{2-\sqrt{3}}$$

Jul 21, 2024