I have a quadrilateral ABCD. If I know the angles $$DAB, DCB, ADB, CDB, ABD, CBD$$, can I obtain the angle $$DAC$$? I know that I can do it by some coordinate geometry or the sine formula, but is there a more elegant way?

• Can you explicitly show what you already know about how to solve this problem ? Commented May 16, 2022 at 9:17

If you need general answer for any values of angles there is no way to avoid trigonometric functions.

Let $$\angle ADB=\alpha$$, $$\angle ABD=\beta$$, $$\angle CDB=\gamma$$, $$\angle CBD=\delta$$, $$\angle DAC=x$$.

Then $$\angle DAB=\pi-(\alpha+\beta)$$, $$\angle DCA=\pi-(\alpha+\gamma+x)$$, $$\angle DCB=\pi-(\gamma+\delta)$$.

Let mark for convenience $$AD=a\sin\beta \sin(\gamma+\delta)$$. Then using sine law:

$$BD=a\sin(\alpha+\beta)\sin(\gamma+\delta), CD=a\sin\delta \sin(\alpha+\beta)$$ $$AD\sin DAC=CD\sin DCA\Rightarrow \sin\beta \sin(\gamma+\delta)\sin x=\sin\delta \sin(\alpha+\beta)\sin(\alpha+\gamma+x)$$ $$\sin\beta\sin(\gamma+\delta)\sin x=\sin\delta \sin(\alpha+\beta)(\sin(\alpha+\gamma)\cos x+\cos(\alpha+\gamma)\sin x)$$ $$\tan x=\frac{\sin\delta \sin(\alpha+\beta)\sin(\alpha+\gamma)}{\sin\beta\sin(\gamma+\delta)-\sin\delta \sin(\alpha+\beta)\cos(\alpha+\gamma)}$$