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#### Help seeker

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Sector AOC is 150/360= 15/36= 5/12 of the entire circle, The entire circle has area $\pi r^2$ so sector AOC is $5\pi r^2/12$. Triangle AOC is an isosceles triangle with two sides of length r and vertex angle 150 degrees. An altitude of triangle AOC bisects angle AOC so divides triangle AOC into two right angles with hypotenuse of length r and one angle 150/= 75 degrees. The "near side", the altitude, has length h= r cos(75) and the "opposite side", half of AC, has length b= r sin(75). All of AC, the base of triangle AOC, has length 2r sin(75) so triangle (1/2)hb= r^2 sin(75)cos(75).

So the shaded segment has area $\frac{5\pi r^2}{12}- r^2sin(75)cos(7)= \left(\frac{5\pi r^2}{12}- sin(75)cos(75)\right)r^2$.

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Tnx.

__SOLVED__

- Feb 21, 2015

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Minor typo....

So the shaded segment has area \(\displaystyle \frac{5\pi r^2}{12}- r^2sin(75)cos(7)= \left(\frac{5\pi r^2}{12}- sin(75)cos(75)\right)r^2\)

That should have been

\(\displaystyle \frac{5\pi r^2}{12}- r^2sin(75)cos(75)= \left(\frac{5\pi}{12}- sin(75)cos(75)\right)r^2\)