Re: Salim's question at Yahoo! Answers regarding trignometry and circular sectors
Hello Salim,
Please refer to the following diagram:
View attachment 1805
We may determine $\theta$ using the Law of Cosines:
$$a^2=2r^2\left(1-\cos(\theta) \right)$$
$$\theta=\cos^{-1}\left(\frac{2r^2-a^2}{2r^2} \right)$$
From this, we may determine the area $A_S$ of circular sector $OBC$:
$$A_S=\frac{1}{2}r^2\theta=\frac{1}{2}r^2\cos^{-1}\left(\frac{2r^2-a^2}{2r^2} \right)$$
And we may now also determine the area $A_T$ of triangle $OBC$:
$$A_T=\frac{1}{2}r^2\sin(\theta)=\frac{1}{2}r^2\frac{a\sqrt{4r^2-a^2}}{2r^2}=\frac{a\sqrt{4r^2-a^2}}{4}$$
Thus the portion of the circle's area $A_O$ outside the triangle is:
$$A_O=\pi r^2-\frac{1}{2}r^2\cos^{-1}\left(\frac{2r^2-a^2}{2r^2} \right)+\frac{a\sqrt{4r^2-a^2}}{4}$$
Equating this to the area of triangle $ABC$, we obtain:
$$\frac{1}{2}bc\sin(A)=\pi r^2-\frac{1}{2}r^2\cos^{-1}\left(\frac{2r^2-a^2}{2r^2} \right)+\frac{a\sqrt{4r^2-a^2}}{4}$$
Hence, solving for $A$, we obtain:
$$A=\sin^{-1}\left(\frac{2r^2\left(2\pi-\cos^{-1}\left(\dfrac{2r^2-a^2}{2r^2} \right) \right)+a\sqrt{4r^2-a^2}}{2bc} \right)$$
