Given that
$$\hat{ABC} = 26^\circ, \hat{ACB} = 52^\circ, |AC| = 11, [AB] \perp [AD] $$
Evaluate $|BD| = x $.
I want to change my point of view against these type of questions. What if we consider that this is a circle? Below I drew a diagram
Is it mathematically right to convert triangle it to such a circle like that? If yes, how can we take it from there?
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Knowing $\angle ABD$ and $\angle BAD$ you can solve for $\angle ADB$ $$\angle ADB=180°-26°-90°=64°$$ The angles $\angle ADB$ and $\angle ADC$ are supplementary, so $\angle ADC=116°$. Knowing angles $\angle ADC, \angle ACD$, and the length of $AC$, the law of sines can be used to solve for the length of $AD$. $$\frac{11}{sin(116)}=\frac{AD}{sin(52)}\Rightarrow AD=\frac{11sin(52)}{sin(116)}\approx9,64$$ The law of sines can be used again to find $x$. $$\frac{9.64}{sin(26)}=\frac{x}{sin(90)}\rightarrow x=\frac{9.64sin(90)}{sin(26)}$$ $x$ is 22.
That is how I would do it. Sorry about the format, I am not familiar with the equation makers.
If we want to use the sine law, to get $x$ we need to look at triangle $BAD$. $$\frac x{\sin 90^\circ}=\frac{AD}{\sin 26^\circ}$$ We don't know $AD$, but we can once again apply the sine law in the $DAC$ triangle. $$\frac{AD}{\sin 52^\circ}=\frac{AC}{\sin \angle{ADC}}$$ You know $AC=11$ and you can get the angle $\angle ADC$ since it's a supplement of $\angle ADB$, and you know all angles in that triangle.