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Let $ABC$ be a right angled triangle with $BC = 3, AC = 4$. Let $D$ be a point in the hypotenuse $AB$ such that $\angle{BCD} = 30^\circ$. Find the length of $CD$.

I found $AB = 5$. How do we find $CD$ ?

Thanks in advance.

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  • $\begingroup$ draw a picture :) $\endgroup$ – danimal Apr 20 '15 at 12:34
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Consider the areas of triangles. Let $[ABC]$ be the area of a triangle $ABC$, and let $CD=x$.

Since $[ABC]=[BCD]+[CDA]$, one has $$\frac{3\times 4}{2}=\frac{1}{2}\cdot 3\cdot x\cdot \sin30^\circ+\frac 12\cdot 4\cdot x\cdot \sin 60^\circ.$$

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