Question

please help..
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Answer 1

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

Answer 2

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.