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

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?

• What else is valid for the triangle? – Maria Mazur Nov 28 '18 at 21:18
• observe that $\angle BDA =138^\circ$ – Vasya Nov 28 '18 at 21:31
• You can use Trigonometric form of Ceva's theorem to find $x$. – Muralidharan Nov 28 '18 at 21:43
• note that $180-24-18=138$, thus $\angle ADB=138$. This gives $$\angle ADC=84$$ – clathratus Nov 28 '18 at 21:50
• I want to directly solve for $x$ using Sine law. Can anyone provide an answer if that's possible? – Hamilton Nov 29 '18 at 13:25

This works if you know what $$|BC|$$ and $$|AC|$$ are.
$$\frac{\sin(x+24)}{|BC|}=\frac{\sin48}{|AC|}$$ $$\sin(x+24)=\frac{|BC|\sin48}{|AC|}$$ $$x+24=\arcsin\bigg(\frac{|BC|\sin48}{|AC|}\bigg)$$ $$x=\arcsin\bigg(\frac{|BC|\sin48}{|AC|}\bigg)-24$$
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.
$$\begin{array}{rcr} x + y + 84 & = & 180 \\ x + y + 84 & = & 180 \end{array}$$