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This seems easy. But it isn't. The diameter is given as $16$ and it asks you to find the coordinates of point $9$. It's tempting to say that it is $(4, 4\sqrt{3})$, but that isn't the answer. What the heck am I doing wrong? We can also assume that the area of each segment is equal

In this triangle, the diameter is 16. Find the coordinates of point (9)

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  • $\begingroup$ Can you explain to me how you got that? That is also not an option for an answer. $\endgroup$ – brandon Apr 25 at 22:29
  • $\begingroup$ Welcome to Math.SE. Where is the origin of the coordinates? $\endgroup$ – Ertxiem Apr 25 at 23:01
  • $\begingroup$ The question doesn't give you the origin of the coordinates, which is why it becomes all the more difficult $\endgroup$ – brandon Apr 25 at 23:18
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    $\begingroup$ Can you please write down all text of the question? From what you've shown, I would say that your answer is correct. $\endgroup$ – Ertxiem Apr 25 at 23:29
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$$x=r\cos\theta$$ $$y=r\sin\theta$$

and $$\theta=\frac{\pi}{3}$$

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  • $\begingroup$ Yeah, and that would yield the answer choice that I mentioned, which is not correct. At least it isn't an answer option. $\endgroup$ – brandon Apr 25 at 22:22
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In parametric form, you have $$x=a+r \cos(t) \qquad \text{and} \qquad y=b+r\sin(t)$$ So $$x_9=a+4 \qquad \text{and} \qquad y_9=b+4\sqrt 3$$

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