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A triangle is inscribed in a circle. The vertices of triangle divide the circle into three arcs of length 3, 4 and 5 units, then find the area of triangle.

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    $\begingroup$ Please leave a comment if you are down voting $\endgroup$ – Kumar Apr 6 '14 at 6:38
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    $\begingroup$ I did not downvote, but this is just a homework question, that "does not show any research effort" (see downvote hover text) $\endgroup$ – Bernhard Apr 6 '14 at 8:39
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enter image description here

Study the above diagram which describes the problem.

From the calculations of the arc lengths, we have:

$$3 = rA, \qquad4 = rB, \qquad 5 = rC$$ $$3 + 4 + 5 = rA + rB + rC = r(A + B + C)$$ $$r = \frac{12}{A + B + C} = \frac{6}{\pi}$$

We also know that the angles $A,B,C$ are in proportions corresponding to the respective arc length proportions. So, $$A = \frac{6\pi}{12}, \qquad B = \frac{8\pi}{12}, \qquad C = \frac{10\pi}{12}$$

I believe it is easy to compute the area of the inscribed triangle from here: It is given by

$$\frac{1}{2}r^2\sin A + \frac{1}{2}r^2\sin B + \frac{1}{2}r^2\sin C$$

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The angles are proportinal to these numbers so their measures are 45, 60 and 75 degrees. Now apply the law of sines to get the sides and then, Heron's formula.

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