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Alright, so i've been trying to tackle this problem. The answer to this problem is 40 degrees (3). Now i'm trying to figure out how.

I decided to start with finding the area of the whole circle which is $9\pi$ cm$^2$ since the radius is 3 cm

but I do not how to proceed from there. Any ideas or hints?

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  • $\begingroup$ There is a formula that allows you to compute the shaded sector from the angle LOM and the radius of the circle. $\endgroup$ – Diego Jul 20 '16 at 5:13
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The area of the shaded sector is $\frac{\theta}2 r^2$, where $\theta = \angle LOM$. Hence, $\theta = 4\pi/9$, so $\angle NOM = 5\pi/9$.

Now $\triangle NOM$ is isosceles at $O$, so $2\angle N = \pi - \angle NOM = 4\pi/9$, hence $\angle N = 2\pi/9 = 40^{\circ}$.

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