4
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Try using the best known formula for triangle area. $\endgroup$ Commented Feb 6, 2023 at 18:24
  • $\begingroup$ @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? $\endgroup$
    – Rusurano
    Commented Feb 6, 2023 at 18:41
  • $\begingroup$ After a round of thinking, it gives precisely nothing because what we want to find actually vanishes. $\endgroup$
    – Rusurano
    Commented Feb 6, 2023 at 19:11
  • $\begingroup$ I was thinking $A=bh/2$ $\endgroup$ Commented Feb 6, 2023 at 20:16

2 Answers 2

1
$\begingroup$

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 $\sin \angle B$.

$\endgroup$
1
  • $\begingroup$ Thanks! I've elaborated your solution in my answer, too $\endgroup$
    – Rusurano
    Commented Feb 8, 2023 at 11:04
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .