# Find the area a this shape inscribed in the unit circle

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: $$\frac{33}{145}-\frac{\sqrt{29}}{29}$$ • It seems to me your solution is right. – user376343 Nov 11 '18 at 21:06
• Is it safe to assume that the textbook solution is wrong? – Daniel Oscar Nov 11 '18 at 21:15
• This time it is the yours that is right. – user376343 Nov 11 '18 at 21:48
• What is the result when you make the multiplications? – Moti Nov 12 '18 at 0:16