what is the value of angle A The triangle ABC is random. The line $AD$ is twice big as the line $DC$ ($AD=2*DC$). We know only the two angles that are shown in the picture. What's the value of angle $A$?
 A: If you only want the measure of the angle (and after deduce with a geometric/trigonometric procedure) Geogebra gives the following:

A: assumption $\overline{CD}=1$
point C $(-1\mid 0)$
point D $(0\mid 0)$
point A $(2\mid 0)$
point B $\left(\frac{1+\sqrt3}2\mid\frac{3+\sqrt3}2\right)\quad$ intersection of
$\qquad\qquad y=x\tan {60°}$ and $y=(x+1)\tan{45°}$
angle A =$\tan^{-1}{\frac{\frac{3+\sqrt3}2}{\frac{1+\sqrt3}2-2}}$
=$\tan^{-1}(2+\sqrt3)=75°$
A: In the figure below we have $$\frac ac=\frac{\sin x}{\sin 45^\circ}\qquad(1)$$ Furthermore
$$\frac{z}{\sin 15^\circ}=\frac{a}{\sin 120^\circ}\qquad(2)$$
$$\frac{2z}{\sin (120^\circ-x)}=\frac{c}{\sin 60^\circ}\qquad(3)$$
From $(1),(2),(3)$ we get $$1.46407\sin x=1.7320\cos x+\sin x\iff \tan x=3.73219$$
Thus $$\color{red}{x\approx 75.00053^\circ}$$

A: I think the correct is answer equal to $75°$
I did it by construction
First of all I made triangle $BDC$ with ur provided information and doing some calculations ie ($\measuredangle DBC=15°$, $\measuredangle BDC=120^{\circ}$ ,$\measuredangle BCD=45°$) than I continued line $CD$ till point $A$ with ur information ($2CD=AD$) as I got point $A$ , I joined it with $B$ and calculated $\measuredangle BAD$ which comes up to be $75°$
A: In triangle BDC you can set up the Sine Law: $\frac{DC}{\sin15^{\circ}}=\frac{BD}{\sin45^{\circ}}$ with $DC=1$, I get $BD=2.732$ Now in triangle BDA we will have to use the Law of Cosine because you have 2 sides with an enclosed angle D so calculating AB we get $AB²=2.732²+2²-2*2.732*2*\cos60°$ which gives $AB=\sqrt{6}$. Now apply Law of Sines in triangle BDA to calculate A. You got three sides and an angle. please give it a try
