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So, if you take a rhombus, then call the bottom left hand corner $A$, the top left hand corner $B$, the top right hand corner $C$, and the bottom right hand corner $D$. Then draw the two diagonals.

The angle of the bottom left hand corner is $60°$. The side lengths are $10$ cm. Call the point where the diagonals bisect each other $X$. I have to find the distance of $AX$.

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  • $\begingroup$ Do you know that the diagonals a rhombus bisect vertex angles and are perpendicular to each other? $\endgroup$ – Maesumi Aug 10 '13 at 13:20
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When you are facing a geometry problem you have no idea how to proceed, it is usually useful to draw an accurate picture on paper/computer and look for hints.

I use GeoGebra to generate the picture at end. As you can see, the line BD and AC looks perpendicular to each other. You can look up facts/theorem about rhombus and indeed it is the case. Since triangle $\triangle AXD$ is a right angled triangle and $\measuredangle DAX = \frac12 \measuredangle DAB = 30^{\circ}$, one get:

$$|AX| = |AD| \cos 30^{\circ} = 10 (\frac{\sqrt{3}}{2}) = 5 \sqrt{3}$$.

A rhombus

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