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I'm having real troubles solving this one geometry problem. I've attempted to draw it as best as I could.

Shape $[ABCD]$ is partly inside the Unit Circle. The only other information states that tan$\left(-\theta -\pi \right)=-\frac{2}{5}$. With this I managed to find sin$\left(\theta \right)=\frac{2\sqrt{29}}{29}$ and cos$\left(\theta \right)=\frac{5\sqrt{29}}{29}$.

Having done that, I concluded that side $AD$ must be $2\cdot$ sin$\left(\theta \right)$ and side $BC$ is $2\cdot$ tan$\left(\theta \right)$. The height is $1-$cos$\left(\theta \right)$. Therefore, the area of $[ABCD]$ is given by: $$\frac{1}{2}\left(\frac{4\sqrt{29}}{29}+\frac{4}{5}\right)\left(1-\frac{5\sqrt{29}}{29}\right)$$ But I must be making a mistake somewhere, since I'm unable to reach the textbook solution by solving that expression, which is: $$\frac{33}{145}-\frac{\sqrt{29}}{29}$$

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  • $\begingroup$ It seems to me your solution is right. $\endgroup$ – user376343 Nov 11 '18 at 21:06
  • $\begingroup$ Is it safe to assume that the textbook solution is wrong? $\endgroup$ – Daniel Oscar Nov 11 '18 at 21:15
  • $\begingroup$ This time it is the yours that is right. $\endgroup$ – user376343 Nov 11 '18 at 21:48
  • $\begingroup$ What is the result when you make the multiplications? $\endgroup$ – Moti Nov 12 '18 at 0:16

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