Revisiting familiar triangles

Let us represent $\cos x$ by $t$ and $\sin x$ by $s$ . Of course, $t^2+s^2=1$ , $t>0, s>0$ , and the given equation becomes $\sqrt{5-4 t} + \sqrt{13-12 s} = \sqrt{10} \:\:\:\:\:(*)$ Squaring both sides of the equation we get $5-4 t+2*\sqrt{5-4 t} \sqrt{13-12 s} +13-12 s=10$ Isolating the term containing the radicals we have $\sqrt{5-4 t} \sqrt{13-12 s} =-4+2 t+6 s.$ Now we square both sides again $(5-4 t) (13-12 s) =16+4 t^{2}+36 s^{2}-48s -16t +24 ts.$ Performing the product in the left hand side and moving all terms to the left $-4t^{2}-36 s^{2} +49 -12 s -36 t+24ts=0.$ Then, replacing $t^{2}$ by $1-s^{2}$ and $s^{2}$ by $1-t^{2}$ we obtain $4s^{2}+36 t^{2} +9 -12 s -36 t+24ts=0.$ Factoring the left side the equation becomes $(2s+6t-3)^{2}=0.$ and, therefore $2s+6t-3 =0$ or, equivalently, $2s= 3-6t\:\:\:\:(**).$ Squaring both sides, replacing $s^{2}$ by $1-t^{2}$ , and moving all terms to the left hand side we obtain $40t^2-36t +5=0.$ with solutions: $t=\frac{9\pm\sqrt{31}}{20}$ . But only one of these values of $t$ is going to produce a positive value of $s$ using $(**)$ . The corresponding values of $t$ and $s$ are going to be $t=\frac{9-\sqrt{31}}{20}$ and $s=\frac{3+3\sqrt{31}}{20}$ . It can be verified by direct calculation that the corresponding pair is the only solution of the equation $(*)$ formed by positive numbers such that the sum of their squares is 1.

Thus the only positive acute angle $x$ for which $\cos x$ and $\sin x$ take, respectively, the previous two values is $x=\cos^{-1} \frac{9-\sqrt{31}}{20}$ , and then the answer for the question is $9+31+20 =60.$

Well, what I thought of this problem is simply to put it into a geometry drawing (although it's one of the subject I would be weak on math...) Notice that sqrt(5 - 4 cos x) = sqrt (2^2 + 1^2 - 2 2 1 cos x) thus first we draw a triangle of side length 1 and 2, with the angle between them as x. Next, sin x = cos (90 - x) so sqrt (13 - 12 sin x) = sqrt (3^2 + 2^2 - 2 3*2 cos (90 - x) hence we draw a triangle of side length 3 and 2 with angle between them as 90 - x, in degrees. We can simply combine the 2 triangles mentioned in one drawing, thus we draw 3 segments, each of length 1, 2, 3, with right angle (90 degrees) between the one of length 1 and 3 and x degree angle between the one of length 1 and 2. Now, let AB be the one with length 1, AC be the one with length 3, and AD be the one with length 2. Then, the equation above is equivalent to BD + CD = sqrt (10). But, BC = sqrt (AB^2 + AC^2) = sqrt (1^2 + 3^2) = sqrt (10), so BD + CD = BC. As a result, D lies on segment BC. Now, we can put it on the Cartesian Coordinate, with the point A (0,0), B (1,0), C(0,3) and we now search for the coordinate of point D. The segment BC follow the line equation: 3x + y = 3 (do I need to explain how we get the line equation?), and since AD = 2, then D is the point (a,b) where 3a + b = 3 and a^2 + b^2 = 4. Notice that now a/2 = cos x. So, substitute b = sqrt (4 - a^2) to the first equation. We get sqrt (4 - a^2) = 3 - 3a. Squaring the equation and then bringing it to one side give us 10a^2 - 18a + 5 = 0. Solving the equation and noticing a =< 2 gives us a = [9 - sqrt (31)]/10. Hence, cos x = a/2 = [9 - sqrt(31)]/20 and thus we get 9+31+20 = 60. Note: you can ask about any unclear part of the solution, but probably I won't be there to make it clear. I'm pretty much sure someone can rebuild my solution in better form...