Parallelogram and side lengths

Question

Using the diagram, find $x$ and $y$ if $ABCD$ is a parallelogram.

Firstly, we can conclude that $$\measuredangle BAC=\measuredangle ACD=25^\circ \text{ (alternate angles)}$$ Then in triangle $ACD$ we will have $\measuredangle CAD=180^\circ-130^\circ-25^\circ=25^\circ\Rightarrow AD=CD, x=y$. The most obvious thing for me now was to apply the Cosine Rule for $\triangle ACD:$ $$AC^2=AD^2+CD^2-2AD\cdot CD\cos130^\circ\\20^2=2x^2-2x^2\cdot\cos130^\circ$$ Well, $130^\circ$ isn't a "tabular" angle, and we haven't studied how to solve such equations. WolframApha says that $x\approx 11.034$. The next thing I tried: Let $DH\perp AC (H\in AC)$. Then $AH=\dfrac12AC=10$ and in $\triangle AHD$ $$\cos25^\circ=\dfrac{AH}{AD}=\dfrac{10}{x}\Rightarrow x=\dfrac{10}{\cos25^\circ}$$ The given answer in my book is $x=y=10\cos25^\circ$. Am I wrong? Thank you!

Answer 1

An idea you might use in the future:

When you said "the most obvious thing ..." it would actually have been been simpler to recognise that, since ACD is isosceles, it splits into two right-angled triangles. Then you immediately have

$$\cos25^\circ=\dfrac{10}{x}. $$

Answer 2

The long answer (Same method):

In $\triangle ABC$
$\angle ABC = \angle ADC = 130°$ (Opposite angles of parallelogram are equal)
$\angle CAB + \angle ACB + \angle ABC = 180°$
$\therefore \angle ACB = 25°$
As 2 angles are same in $\triangle ABC$, it will be an isosceles triangle.
$\therefore AB = BC$

CD = y
AD = x
$\therefore AB = y$ and $BC = x$ (Opposite sides of parallelogram are equal)
But AB = BC
$\therefore AB = BC = CD = AD = x$
$\therefore ABCD$ is a rhombus (Parallelogram with equal sides is a rhombus).

Join BD
In rhombus, the diagonals bisect each other at 90° angle.
Let the diagonals intersect at point O.

In $\triangle COB$
$\angle COB = 90°$
$\sin 65 = \frac{10}{x}$
$\therefore x = \frac{10}{\sin 65}$