Find possible areas of triangle given radius of circumscribed circle

Question

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?

Answer 1

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 $\approx 180-50$.

Answer 2

$AB$ 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.

Answer 3

You'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.