A quadrilateral $ABCD$ as $\angle A = 60^\circ, \angle D = 70^\circ, AB = 10$, and $ BC = CD = DA = x $ where $x$ is unknown. Find $x$ and $\angle C$.
My attempt:
Using the law of cosines on $\triangle ABD $ and $\triangle BCD$, we get
$x^2 + 100 – 20 x \cos 60^\circ = 2 x^2 – 2 x^2 \cos(C) $
And applying the law of cosines on $\triangle ADC $ and $\triangle ABC $, we get
$100 + x^2 – 20 x \cos (230^\circ – C) = 2 x^2 – 2 x^2 \cos(70)$
However, I don't know how to solve these two equations for $x$ and $C$.
Best Answer
As $DAC$ is isoceles, $\angle DAC=55°$. So by cosines law, $$x^2+AC^2-2xAC\cos 55°=x^2\implies AC=2x\cos 55°$$ $\angle CAB=5°$ again by law of cosines, $$AC^2+100-20AC\cos 5°=x^2\\\implies 4x^2\cos^255°+100-40x\cos 55°\cos 5° =x^2$$ This gives $x=67.66,4.6778$
Now $$AC^2+x^2-2ACx\cos ACB=100$$ Substituting values of $x$, We get $ACB=0.737°$,or $ACB=169.25°$
We just add $55°$ to it, which gives us $\angle C=55.737°,224.25°$