I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$

The counterpart questions for sine and tangent can be handled as follows:

  • If $\dfrac{\sin A}{2}=\dfrac{\sin B}{3}=\dfrac{\sin C}{7}$, we can rule out triangle because by the Sine Rule $a=2k$, $b=3k$, $c=7k \implies a+b <c.$

  • If $\dfrac{\tan A}{2}=\dfrac{\tan B}{3}=\dfrac{\tan C}{7}$, we can see that a triangle will be made as $\tan A=2k, \tan B =3k,\tan C=7k$, when inserted in the identity $\tan A+ \tan B+ \tan C= \tan A \tan B \tan C \implies k=\sqrt{2/7}$.

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    $\begingroup$ @LionHeart: Yes, Dharmendra knows this. Read the question again. $\endgroup$
    – TonyK
    Sep 21, 2020 at 10:51

3 Answers 3


Show for yourself that if $A,B,C$ are the angles of a triangle then $$ \cos^2A + \cos^2B + \cos^2C = 1-2\cos A \cos B \cos C $$

This is not very difficult, use the fact that $A+B+C = 180$ along with double angle identities.

Therefore, if $$ \frac{\cos A}{2} = \frac{\cos B}{3} = \frac{\cos C}{7} = k, $$ then by substituting this in the formula in the first section, $$62k^2 = 1-84k^3.$$ This can be solved using Cardano's formula (or you can use the intermediate value theorem to assert the existence of a root in $[0,1]$). Numerical methods show that $k \approx 0.117928$. From here, you get that such a triangle in fact exists, and roughly has angles $76.36^\circ,69.28^\circ$ and $34.36^\circ$.

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    $\begingroup$ I mentioned this : you need not solve the cubic, but only need to assert the existence of a root between $k=0$ and $k = \frac 17$ so that each of the cosines are well defined. The cubic has opposite signs for $k=0$ and $k=\frac 17$ (the latter requires some hard work, but can be achieved), then by Bolzano's Theorem, at some point of time between these points the cubic is $0$. I used Wolfram to find that value, otherwise unfortunately you will need to use Cardano's formula and solve the cubic if you want the exact value. $\endgroup$ Sep 21, 2020 at 11:09
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    $\begingroup$ It is worth mentioning that $k \approx -0.141$ is also a root, but $A + B + C \ne \pi$ for this value of $k$. $\endgroup$
    – Toby Mak
    Sep 21, 2020 at 11:12
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    $\begingroup$ @TobyMak Thank you. Indeed, $k<0$ then each of the cosines are negative, but then each of $A,B,C$ have to be obtuse, which is not possible. $\endgroup$ Sep 21, 2020 at 11:17
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    $\begingroup$ It seems to me that the angles $77.3,70.7 $ and $39$ are too rough since $77.3+70.7+39=187$. $\endgroup$
    – mathlove
    Sep 21, 2020 at 11:20
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    $\begingroup$ $f(k)=84k^3+62k^2-1 \implies f(0)=-1, f(1/7)=\frac{25}{49}>0$ so the equation has at least one root in (0,1/7). This also is very importan for the existence of the triangle. $\endgroup$
    – Z Ahmed
    Sep 21, 2020 at 18:01

Strating from @Teresa Lisbon's answer, the exact results are $$k=\frac{31}{126} \left(2 \cos \left(\frac{1}{3} \left(2 \pi n-\cos ^{-1}\left(-\frac{17884}{29791}\right)\right)\right)-1\right)\qquad (n=0,1,2)$$ and this gives angles (in degrees) $a=76.358$, $b=69.281$, $c=34.361$.

Using algebra, the problem is very simple since it reduces to the equation $$a+\cos ^{-1}\left(\frac{3}{2} \cos (a)\right)+\cos ^{-1}\left(\frac{7 }{2} \cos (a)\right)=\pi$$ which has only one solution.

Using series expansion around $a=\frac \pi 2$ gives $$0=\frac{\pi }{2}+6 \left(a-\frac{\pi }{2}\right)+\frac{55}{8} \left(a-\frac{\pi }{2}\right)^3+\frac{4627}{128} \left(a-\frac{\pi }{2}\right)^5+O\left(\left(a-\frac{\pi }{2}\right)^7\right)$$ and series reversion leads to $$a \sim\frac{5 \pi }{12}+\frac{55 \pi ^3}{82944}+\frac{89 \pi ^5}{10616832}\approx 1.33212$$ while the "exact" solution is $a=1.33270$.


Hint use may use the identity

$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$$


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