How do we solve for $x$ using law of sines?

Question

How do we solve for $x$ using the law of sines?

Here's my attempt:

$$\angle BDC = 180- 30 - 12 = 138^\circ$$

So we have that

$$\dfrac{\sin 30 }{|DC|} = \dfrac{\sin 12}{|DB|} = \dfrac{\sin 138}{|BC|} $$

For $\triangle{ABD}$

$$\dfrac{\sin 18 }{|AD|} = \dfrac{\sin 24}{|DB|}$$

And what I need to evaluate is

$$\dfrac{\sin x }{|DC|} = ? $$

Can you help me take it from here?

Answer 1

This works if you know what $|BC|$ and $|AC|$ are.

$$\frac{\sin(x+24)}{|BC|}=\frac{\sin48}{|AC|}$$ $$\sin(x+24)=\frac{|BC|\sin48}{|AC|}$$ $$x+24=\arcsin\bigg(\frac{|BC|\sin48}{|AC|}\bigg)$$ $$x=\arcsin\bigg(\frac{|BC|\sin48}{|AC|}\bigg)-24$$

Answer 2

Recall that the angles of a triangle must add up to 180°. From this we can conclude that angle $CDB$ measures 138° and angle $BDA$ measures 138°. Therefore angle $ADC$ measures 84°. Now, if we let $y$ be the measure of angle $DCA$ (in degrees), we have the following system of equations.

$$ \begin{array}{rcr} x + y + 84 & = & 180 \\ x + y + 84 & = & 180 \end{array} $$

So I want to say there will be infinitely many solutions... apparently there is some freedom in choosing the sides, as the other answer has hinted. But not 100% sure. Anyone else want to weigh in?