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Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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work the area of the triangle as \((1/2 * base * height) * 2\) \(\large = \frac{1}{2}. r cos(\frac{\pi - x}{2}).r sin(\frac{\pi - x}{2}) \times 2\) using double angle formula, add all 3 areas up to get total \( \frac{\pi r^2}{2} \) and it falls out of that
Easy! Find the area of the triangle AOC. As AO = OC = r, the angles OAC and OCA are = x / 2. Let the length of AC be d and the length of the perpendicular from O on AC be h. Now, \[arc segment AOC - \triangle AOC = 1/ 4 \times (\pi r^2 / 2)\] Now the arc segment is \[r^2 / 2 \times (\pi - x)\] Use sine law for the triangle to get \[\triangle AOC = \frac{1}{2} hd = 1/2 \times r^2 \sin x\] Now solve the equation to get the answer.

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