Question

A rhombus ABCD has AB = 10 and m∠A = 60°. Find the lengths of the diagonals of ABCD.

Answer 1

Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus

First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°. 

Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.

For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.

$\overline{OB}:\overline{AB} = 1:2$
$\overline {OB}:10 = 1:2$
$\overline{OB} = \frac{1}{2}(10) = 5$

Similarly, we have

$\overline{AO}:\overline{AB} = \sqrt{3}:2$
$\overline {AO}:10 = \sqrt{3}:2$
$\overline{AO} = \frac{\sqrt{3}}{2}(10) = 5\sqrt{3}$

Now, to find the lengths of the diagonals, 

$\overline{AD} = 2(\overline{AO}) = 10\sqrt{3}$
$\overline{BC} = 2(\overline{OB}) = 10$

So, the lengths of the diagonals are 10 and 10√3.

Answer: 10 and 10√3 units