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# Help appreciated. Thanks!

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The vertices of triangle$$ABC$$ lie on the sides of equilateral triangle$$DEF$$ , as shown. If $$CD=3$$, $$CE=BD=2$$, and $$\angle ACB =90^{\circ}$$, then find $$AE$$.

Mar 16, 2022

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$$Find\ \overline{AE}.$$

Hello Guest!

$$\overline{BC}=\sqrt{2^2+3^2-2\cdot 3\cdot cos\ 60°}\\ \overline{BC}=\sqrt{10}$$

$$\frac{sin\ BCD}{2}=\frac{sin\ 60°}{\sqrt{10}}\\ sin\ BCD=\frac{2\cdot sin\ 60°}{\sqrt{10}}\\ BCD=33.211°$$

$$ACE=180°-90°-33.211°\\ ACE=56.789°$$

$$EAC=180°-60°-56.789°\\ EAC=63.211°$$

$$\frac{\overline{AE}}{sin\ 56.789°}=\frac{2}{sin\ 63.211°}\\ \overline{AE}=\frac{2\cdot sin\ 56.789°}{sin\ 63.211°}$$

$$\overline{AE}=1.8745$$

!

Mar 16, 2022