# System the system with $3\sin a+4 \cos b = 6$ and $4 \sin b+3 \cos a=1$

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?

• 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. May 16, 2019 at 17:49
• Edited the last sentence to make that clear. It doesn't matter much, since the accepted answer implicitly answers that as well. May 16, 2019 at 18:03

Hint

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

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

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

$$\iff8\sqrt{6^2+1^2}\sin(B+\arcsin\dfrac6{\sqrt{37}})=44$$

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

Similarly for $$A$$

• 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. May 16, 2019 at 18:18
• Aw.. This was my strategy except mine is unnecessarily long and I just finished. I guess I won't post it. +1 May 16, 2019 at 18:21
• Hmmm.. .however, I did do my for $a$ so maybe I will. 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:

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 $$b=\frac{\pi}{2}+\arctan\left(\frac{1}{39}(11\sqrt{3}-8)\right)\approx 105^\circ 49'22''$$.

$$\cos^2(b)+\sin^2(b)=1$$

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

$$\cos(b)=\frac{6}{4}-\frac{3}{4}\sin(a)$$

and

$$\sin(b)=\frac{1}{4}-\frac{3}{4}\cos(a)$$

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:

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

Let's simplify a bit

$$\frac{60}{\sqrt{5328}}=\sin(a-\theta)$$

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