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Question: In triangle ABC, AB=4, AC=5, and the radius R of the circumscribed circle is equal to √7. Find all possible values of the area of triangle ABC.

Through using the sine rule, i found the angle of B to be approximately $70.89339$, the angle of C to be approximately $49.10661$ and the angle of A to be $60$. Using the cosine law, I found the missing side length BC to be $√21$. Using Heron's law, I then found the area to be $5√3$.

However, this question requires more than one area. I am confused as to how to obtain the second area?

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$AB$ and $AC$ are two chords of a circle of radius $\sqrt7$. Picture this circle together with its diameter through $A$. The two chords can either be on the same side or opposite sides of this radius. You’ve computed the area of the resulting triangle for only one of these cases.

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Remember that the sine rule cannot tell the difference between $\theta$ and $180-\theta$. Hence you have to check for the possibility of having an obtuse angle. If the angle at $B$ is $\approx 180-71 = 109$ it still leaves room for the angle at $C$ to be $\approx 50$. This will give a different area.

You should also check what happens when you replace $\approx 50$ by $\approx 180-50$.

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enter image description hereYou're right, there is only one possible answer. Since $M$ is the circumcenter of $\triangle ABC $, $MD $ is the perpendicular bisector of $\overline{AB} $. Applying the sine law in $\triangle ABC $ then gives $$|\measuredangle AMD|=\arcsin (\frac {2}{\sqrt {7}}). $$ Analogously, we obtain $$|\measuredangle EMD|=\arcsin (\frac {2.5}{\sqrt {7}}). $$ Overall this leads to $|\measuredangle EMD|= \frac {2\pi}{3} $ and therefore we get $|\measuredangle BAC|=|\measuredangle DAE|=\frac {\pi}{3 } $. Hence, the area of our triangle comes out to be $\frac {1}{2} \cdot 4 \cdot 5 \cdot \sin (\frac {\pi}{3 }) =5\sqrt {3}$, which is exactly your result.

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    $\begingroup$ There is another possibility: the point $C$ can be on the same side of the diameter through $A$ and $B$ instead of on the opposite side as you’ve drawn it. $\endgroup$ – amd Apr 12 '17 at 0:28
  • $\begingroup$ @amd This cannot be the case since BAC would be acute then but also be equal to $arcsin (\frac {2}{\sqrt{7}})-arcsin (\frac {2.5}{\sqrt{7}})$. But this is roundabout $-22$ degrees or about $158$ degrees which cannot be right in both cases. Hence, what I've drawn is the only case that can occur. $\endgroup$ – mxian Apr 12 '17 at 0:37
  • $\begingroup$ Draw a circle of radius $5$ centered at $A$. It intersects the circle that you’ve drawn at two points, both of which are perfectly valid choices for $C$. Or, simply reflect the point $C$ that you’ve drawn in the diameter through $A$ to see that there’s another possible triangle. Whoever put together the problem that the OP is trying to solve certainly believed that there’s more than one non-congruent possibility for the inscribed triangle. $\endgroup$ – amd Apr 12 '17 at 1:11

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