You have these two equations (a, b, c are angles of a triangle): $$ 3\sin(a)+4\cos(b)=6\\ 4\sin(b)+3\cos(a)=1 $$

The triangle part limits a, b, c to (0, $\pi$) and provides an additional equation:

$$ a+b+c=\pi $$

You are supposed to figure out c. There are two solutions, $\frac{\pi}{6}$ and $\frac{5\pi}{6}$, but $\frac{5\pi}{6}$ has no corresponding triangle.

However, there's no obvious proof that the first solution does have a corresponding triangle. Wolfram Alpha did find a solution, but not the step-by-step guide.

How do you find a and b?

As a secondary question, the first equation, combined with the value of c (which makes b a function of a), only gives a very narrow range of possible values of a and b. For these values, there happens to exist a solution of the second equation. If the 1 was changed to a different number, this would seem to not be the case, but for that case, there's a different result of c.

So, how does the equations' solvability depend on the numbers on the right side?

  • $\begingroup$ Which is exactly the question?! ...of course the solvability depends on the numbers $6,1$ on the R.H.S of the given equations, taking $6000, 1$ leads to no solution. $\endgroup$
    – dan_fulea
    May 16, 2019 at 17:49
  • $\begingroup$ Edited the last sentence to make that clear. It doesn't matter much, since the accepted answer implicitly answers that as well. $\endgroup$
    – Kotlopou
    May 16, 2019 at 18:03

3 Answers 3



$$(3\cos A)^2=(1-4\sin B)^2$$

$$(3\sin A)^2=(6-4\cos B)^2$$

Adding we get $$9=37+16-8(\sin B+6\cos B)$$


$$B=\pi-\arcsin\dfrac{11}{2\sqrt{37}}-\arcsin\dfrac6{\sqrt{37}}$$ as $B>0$

Similarly for $A$

  • $\begingroup$ According to Wolframalpha, the $B$ that you found simplifies to $$B = \arcsin\left(\dfrac{11+18\sqrt{3}}{74}\right)$$ but I am not seeing how... Not really necessary, but looked like a nice simplification if I could figure it out. $\endgroup$ May 16, 2019 at 18:18
  • $\begingroup$ Aw.. This was my strategy except mine is unnecessarily long and I just finished. I guess I won't post it. +1 $\endgroup$
    – randomgirl
    May 16, 2019 at 18:21
  • $\begingroup$ Hmmm.. .however, I did do my for $a$ so maybe I will. $\endgroup$
    – randomgirl
    May 16, 2019 at 18:23

The whole system can be written as $3e^{ia}+4e^{ib}=1+6i$. It can be solved by finding the angles of a triangle whose side lengths are $3,4$ and $\sqrt{37}=\left|1+6i\right|$. Geometrically:

enter image description here

By intersecting a circle with radius $4$ centered at $(1,6)$ and a circle with radius $3$ centered at the origin we get two solutions, $$ (x,y) = \left(\frac{3}{37}(5\mp 12\sqrt{3}),\frac{6}{37}(15\pm\sqrt{3})\right) $$ one of them being associated to $a=\arctan\left(\frac{2}{11}(5\sqrt{3}-3)\right)\approx 45^\circ 49'22''$ and $b=\frac{\pi}{2}+\arctan\left(\frac{1}{39}(11\sqrt{3}-8)\right)\approx 105^\circ 49'22''$.



Solving both equations for the trig(b) parts, that is, this:




Applying the Pythagorean Identity in first line here gives:

$1=\frac{37}{16}-\frac{9}{4} \sin(a)-\frac{3}{8} \cos(a)+\frac{9}{16}$

So rewriting so just the trig functions are one side gives:

$\frac{15}{8}=\frac{9}{4} \sin(a)+\frac{3}{8} \cos(a)$

Stop for a second with that. Let's think about right triangles now. Let's find a triangle whose legs are 9/4 units and 3/8 units. The hypotenuse would be $\sqrt{(\frac{9}{4})^2+(\frac{3}{8})^2}=\frac{\sqrt{5328}}{32}$.

So we are going to divide both sides by that!

$\frac{\frac{15}{8}}{\frac{\sqrt{5328}}{32}}=\frac{\frac{9}{4}}{\frac{\sqrt{5328}}{32}} \sin(a)+\frac{\frac{3}{8}}{\frac{\sqrt{5328}}{32}} \cos(a)$

Let $\cos(\theta)=\frac{\frac{9}{4}}{\frac{\sqrt{5328}}{32}}$ and $\sin(\theta)=\frac{\frac{3}{8}}{\frac{\sqrt{5328}}{32}}$.

So now we can rewrite this as:

$\frac{\frac{15}{8}}{\frac{\sqrt{5328}}{32}}=\cos(\theta) \sin(a)+\sin(\theta) \cos(a)$

Which can then be rewritten as:


Let's simplify a bit


We can use this to find $a$. Then in a similar way we can find $b$.


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