# Solving a triangle by angle, circumradius, and area

The problem itself is as follows.

Consider a triangle $$ABC$$. The cosine of one of its interior angles is $$m$$, its circumradius is equal to $$R$$, and its area is equal to $$S$$. Solve this triangle: find its sides and angles.

By applying the sine theorem, I've got that the length of a side, opposite to the angle with the known cosine, which is $$2R\sqrt{1-m^2}$$. But I had no luck in advancing further, as it either results in too many unknowns. Most I could do is getting the product of sines of two unknown angles, from the formula $$S=2R^2\sin\angle A\sin\angle B\sin\angle C$$.

I'm looking for proofs that such triangle exists and is clearly defined, as well as the way to solve it if it is.

• Try using the best known formula for triangle area. Commented Feb 6, 2023 at 18:24
• @AaronGoldsmith Half the product of two sides and sine of the angle between these sides? I've tried using it, but isn't $0.25abc/R = 0.5ab\sin\angle C$ resulting in too many unknowns? Commented Feb 6, 2023 at 18:41
• After a round of thinking, it gives precisely nothing because what we want to find actually vanishes. Commented Feb 6, 2023 at 19:11
• I was thinking $A=bh/2$ Commented Feb 6, 2023 at 20:16

This is a sketch of a solution. Let's say wlog that $$\cos \angle A=m$$. We have $$a=2R\sqrt{1-m^2}$$. We also have $$2A=bc \sqrt{1-m^2}$$ and $$b=2R\sin \angle B$$, $$c=2R\sin \angle C$$. On the other hand, $$a^2=c^2+b^2-2bc \cdot m$$.

Thus, $$4R^2(1-m^2)=4R^2\sin^2 \angle B+4R^2\sin^2 \angle C-\frac{4A\cdot m}{\sqrt{1-m^2}}$$.
Finally, $$\sin \angle C=\sin (\angle A+\angle B)=m\sin \angle B+\sqrt{1-m^2}\cos \angle B$$. Substituting this into above will give us equation to find $$\sin \angle B$$.

• Thanks! I've elaborated your solution in my answer, too Commented Feb 8, 2023 at 11:04

Below is my own answer on the problem I came up with. There may exist more elegant ways, but this is the most straightforward way of which I know.

Consider a triangle $$ABC$$ with circumradius $$R$$, area $$S$$, and $$\cos\angle C=m$$.

1. Find the sine of $$\angle C$$ by using the famous trigonometric identity $$\sin^2\alpha+\cos^2\alpha=1$$. As all interior angles of the triangle have positive sines, we take the positive value, thus $$\sin\angle C=\sqrt{1-\cos^2\angle C}=\sqrt{1-m^2}$$.
2. Find the length of side $$AB$$, opposite to $$\angle C$$, by using the law of sines. Thus, $$AB=2R\sin\angle C=2R\sqrt{1-m^2}$$.
3. Find the product of two other sides. For that, find the altitude of triangle $$ABC$$ to side $$AB$$ by using the formula $$h_{AB}=\dfrac{2S}{AB}$$. Now, if we equate two formulas for the area of a triangle, we get $$\dfrac{AB\cdot BC\cdot AC}{4R}=\dfrac{1}{2}AB\cdot h_{AB}$$, which we can transform further: \begin{align*} \dfrac{AB\cdot BC\cdot AC}{4R} = \dfrac{1}{2}AB\cdot h_{AB}\ \ &\Leftrightarrow\ \ \dfrac{BC\cdot AC}{2R} = h_{AB} \\ &\Leftrightarrow\ \ \dfrac{BC\cdot AC}{2R} = \dfrac{2S}{AB} \\ &\Leftrightarrow\ \ BC\cdot AC = \dfrac{4RS}{AB} \end{align*} Thus, $$BC = \dfrac{4RS}{AB\cdot AC}$$.
4. By using the law of cosines, we get $$AB^2=BC^2+AC^2-2\cdot BC\cdot AC\cos\angle C$$. Here, we can substitute $$BC\cdot AC = \dfrac{4RS}{AB}$$, and $$BC = \dfrac{4RS}{AB\cdot AC}$$. We get: \begin{align*} AB^2=\left(\dfrac{4RS}{AB\cdot AC}\right)^2+AC^2-2\cdot \dfrac{4mRS}{AB} \end{align*}

Here, we know everything except $$AC$$, and we can find both unknown sides by simplifying the last formula into the biquadratic equation, positive roots of which are sides $$BC$$ and $$AC$$. Finally, to find the other angles, one can use the law of cosines if the given angle is obtuse ($$m < 0$$), as there is only one obtuse angle in the triangle, or the law of cosines in all other cases.