1. anonymous

2. IrishBoy123

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

3. anonymous

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.