# Finding half a diagonal in a rhombus using trigonometric ratios?

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 $AX$.

• Do you know that the diagonals a rhombus bisect vertex angles and are perpendicular to each other? – Maesumi Aug 10 '13 at 13:20

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