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I started out with a system of equation with 5 variables.

$73 + 34 + x_{3} + x_{5} + x_{2} + 30 + x_{4} - 360$

$73 + x_{4} + 34 - 180$

$-150 + x_{2} + x_{3} + x_{5}$

$73 - x_{1} +34 x_{2} + x_{3}-180$

$73 - x_{1} + 1/2 x_{2} + 90 - 180$

See the picture below. Then I simplify and use matrix to solve the system. It turns out that the system is homogeneous and $x_{5}$ is a free variable. I got lost here and don't know how to proceed. Please give me some help or if you have a better method, please elaborate. thanks.

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Note that $\angle ADB = 73°$ and $\triangle ADB$ is isosceles.

Construct a point $E$ that makes an equilateral triangle with $BC$ towards $A$. Then $D$ is on the circle centred at $E$ through $B$ and $C$ (since $30°$ is half $60°$), so $DE=BE=CE=BC=AC$. Also note that since $A,E,B$ are on a circle centred at $C$, $\angle EAB=30°$ also and $\triangle AED \cong \triangle ADC $.

That gives $CE$ is parallel to $DA$ and thus $\angle DBC = \angle ABE = (60-34)/2 = \fbox{13°}$

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    @ Joffan how do you make an equilateral triangle with BC towards A? – 2017-02-25
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    Also, how do you know D is on the circle centered at E? – 2017-02-25
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    Typo: $DE$ is parallel to $BC$ – 2017-02-25
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    @GNUSupporter no, it was a typo though, so thanks; CE is parallel to DA. – 2017-02-25
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    @Tmm I know that D is on that circle because $\angle BDC $ is half the angle at the centre $\angle BEC$. I can build an equilateral triangle on any line segment, with a choice of which way it is facing, just based on the length of the line. – 2017-02-25
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    @ Joffan I am sorry, I still don't understand how $\angle BDC$ is half the angle at the centre of $\angle BEC$. Would you mind giving me more information? – 2017-02-25
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    @Tmm by construction – 2017-02-25
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    @ GNU Supporter would you elaborate construction? – 2017-02-25
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    $\angle BDC$ is given as $30°$ and $\angle BEC$ is constructed as $60°$. Clearly you can draw a circle with $E$ central and $B,C$ on the circumference, so the half-angle gives a point on that circle https://en.wikipedia.org/wiki/Inscribed_angle – 2017-02-25
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    Also note that $\angle EAB$ is $30°$ from the circle centred on $C$. – 2017-02-25