"Determinate a value for $\angle DBC$ if $\angle DAC = 2\angle DCA= 40º$ and $BC=\sqrt 3\space AD$.
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After trying this problem, i ended with $\sin \angle DBC = \cos 20º$
So this problem has 2 answers: $\angle DBC = 70º$ or $\angle DBC = 110º$ and it seems like both answers are valid.
It's this possible?. Why?
On
It's possible and, in general, there will be multiple solutions to a problem of this form (with varying parameters for the $40^\circ$ and $\sqrt3$ values you've provided). This is because for most lengths $BC$, there are two points $B$ and $B'$ that satisfy the conditions in the problem, as shown below:
Here, $\triangle BB'C$ is meant to be isosceles. Nothing given to us changes when we replace $B$ by $B'$.
We will get a unique solution only when that solution is $\angle ABC = 90^\circ$.
For some versions of this problem, we could possibly eliminate one of the values if it would lead to $B'$ being between $A$ and $D$. However, if your solutions of $70^\circ$ and $110^\circ$ are correct (I haven't checked), then both points are on the same side of $D$ (because $\angle ADC = 120^\circ$ is larger than both $70^\circ$ and $110^\circ$), and this doesn't help you narrow things down here.
You can transform $\sin$ to $\cos$ by writing $\sin x=\cos (90^\circ-x)$. So this would yield $90^\circ-x=20^\circ$ or $x=70^\circ$. But note that cosine is an even function, so $\cos(90^\circ-x)=\cos(x-90^\circ)$. Plugging this in, would result in $90^\circ-x=20^\circ$ or $x=110^\circ$.