Hello, Hanga!

If I know that the sides of a parallellogram is 1 and 2, and that one angle is 60°,

can I, using plain Euclidean Geometry, calculate the diagonals exactly? . . . . Yes!

We have parallelogram ABCD with: .$\displaystyle AB = CD = 2,\;AD = BC = 1,\;\angle B = \angle D = 60^o$

This parallelogram can be divided into four equilateral triangles!

Code:

A * - - - - - * - - - - - * B
/ \ / \ /:
/ \ / \ / : _
1 / \ / \ / : ½√3
/ \ / \ / :
/ 60° \ / \ / :
* - - - - - * - - - - - * - - +
D 1 1 C ½ E

Draw diagonal $\displaystyle AC.$

It can be seen that $\displaystyle \Delta DAC$ is a right triangle

. . with: $\displaystyle \angle DAC = 90^o,\:AD = 1,\: hyp = 2$

Hence: .$\displaystyle AC \,=\,\sqrt{3}$

Draw diagonal $\displaystyle BD.$

In right triangle $\displaystyle BED\!:$

. . $\displaystyle BD^2 \:=\:DE^2 + BE^2 \:=\:\left(\tfrac{5}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 \:=\:\tfrac{25}{4} + \tfrac{3}{4} \:=\:\tfrac{28}{4} \:=\:7$

Therefore: .$\displaystyle BD \:=\:\sqrt{7}$