The triangle ABC is a right angle triangle. I'm trying to figure out the distance AC, i.e the height of the triangle.
My bet is that I could use the uniformity of triangle ABC and CDO to solve the height. Let AC=h and OD=b. That should mean that $$\frac{h-3.5}{h}=\frac{b}{7}=\frac{\sqrt{b^2+(h-3.5)^2}}{\sqrt{7^2+h^2}}$$ However there's a problem. I have to unknown variables so I can't get any further from here. Maybe I can use the fact that I can express b in form of h, such as $$\frac{b}{7}=\frac{h-3.5}{h}=>b=7\frac{h-3.5}{h}$$ I also know that the angle $$\angle AOB=arcsin(\frac{7}{\sqrt{7^2+3.5^2}})=63.43°=>\angle ABO=180°-90°-63.43°=26.57°$$ I feel like if I could figure out the $\angle OBD$, I could calculate the heigh h. However, in order for me to figure out $\angle OBD$, I need to find out DB, which should be $CB-CD$. $$DB=CB-CD=\sqrt{7^2+h^2}-\sqrt{(\frac{7(h-3.5)}{h})^2+(h-3.5)^2 }$$ Therefore the angle $\angle OBD$ could be found from $$\frac{sin(\angle OBD)}{\frac{7(h-3.5)}{h}}=\frac{\angle DOB}{DB}==\frac{26.57°}{\sqrt{7^2+h^2}-\sqrt{(\frac{7(h-3.5)}{h})^2+(h-3.5)^2 }}$$ I think I've either done something wrong because they numbers are getting silly now or I'm using the wrong technique. Does anyone have an idea on how I could calculate the height?
It depends what are the know factors. Usually when you have parallel lines in a triangle, Thales is your friend.
You have $\dfrac{CO}{CA}=\dfrac{OD}{AB}$, which means
$\dfrac{CA-3.5}{CA}=\dfrac{OD}{7}$, or $AC=\dfrac{3.5}{1-\frac{OD}{7}}$