Uhmm, I doubt that your friend is telling the truth. This problem looks horrible.
Now arccos(x) is always equal to or greater than 0, right?, since: 0 \leq \arccos (x) \leq \pi, so that means:
\arctan \left( \frac{8}{x} \right) = \arccos \left( \frac{x + y}{20} \right) \geq 0, hence x > 0, since \arctan \left( \frac{8}{x} \right) \geq 0.
Do the same for the second equation to obtain y > 0.
Let:
\arctan \left( \frac{8}{x} \right) = \arccos \left( \frac{x + y}{20} \right) = \alpha. That means:
\tan \alpha = \frac{8}{x} \Leftrightarrow 1 + \tan ^ 2 \alpha = \frac{x ^ 2 + 64}{x ^ 2} \Leftrightarrow \frac{1}{\cos ^ 2 \alpha} = \frac{x ^ 2 + 64}{x ^ 2} \Leftrightarrow \cos ^ 2 \alpha = \frac{x ^ 2}{x ^ 2 + 64}
And from \arccos \left( \frac{x + y}{20} \right) = \alpha, you'll have: \cos \alpha = \frac{x + y}{20}, so we have:
\frac{(x + y) ^ 2}{400} = \frac{x ^ 2}{x ^ 2 + 64} \quad (1).
Do the same for the second equation to obtain:
\frac{(x + y) ^ 2}{900} = \frac{y ^ 2}{y ^ 2 + 64} \quad (2)
Eleminate (x + y)2 from 2 equations above (Equation (1), and equation (2)) to abtain:
\frac{x ^ 2}{x ^ 2 + 64} = \frac{9}{4} \times \frac{y ^ 2}{y ^ 2 + 64}
Solve for y in terms of x, we have:
y ^ 2 = \frac{256 x ^ 2}{5 x ^ 2 + 576} \quad (3)
Substitute (3) into (1), we have:
\frac{x ^ 2 + \frac{32 x ^ 2}{\sqrt{5x ^ 2 + 576}} + \frac{256 x ^ 2}{5 x ^ 2 + 576}}{400} = \frac{x ^ 2}{x ^ 2 + 64}
\Leftrightarrow \left( x ^ 2 + \frac{32 x ^ 2}{\sqrt{5x ^ 2 + 576}} + \frac{256 x ^ 2}{5 x ^ 2 + 576} \right) (x ^ 2 + 64) = 400 x ^ 2
\Leftrightarrow ( x ^ 2 (5 x ^ 2 + 576) + 32 x ^ 2 \sqrt{5x ^ 2 + 576} + 256 x ^ 2 ) (x ^ 2 + 64) = 400 x ^ 2 (5 x ^ 2 + 576)
\Leftrightarrow ( 5 x ^ 2 + 576 + 32 \sqrt{5x ^ 2 + 576} + 256 ) (x ^ 2 + 64) = 400 (5 x ^ 2 + 576)
\Leftrightarrow (32 \sqrt{5x ^ 2 + 576}(x ^ 2 + 64) = 400 (5 x ^ 2 + 576) - (5x ^ 2 + 832) (x ^ 2 + 64)
From there, square both sides, you'll have a quartic equation in x2
Something like this:
25 x^8-13600 x^6-2297600 x^4+203980800 x^2+28966912000 = 0
And there are 2 real roots:
\left[ \begin{array}{l} x ^ 2 \approx 122.6338468 \\ x ^ 2 \approx 660.42895867809 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x \approx 11.074 \\ x \approx 25.6988 \end{array} \right.
The second x is not valid since \frac{x + y}{20} \approx \frac{25.6988 + y}{20} > 1
From the first x, we have:
y = \sqrt {\frac{256 x ^ 2}{5 x ^ 2 + 576}} = \frac{16 x}{\sqrt{5x ^ 2 + 576}} \approx 5.138
So the solution is:
\left\{ \begin{array}{l} x \approx 11.074 \\ y \approx 5.138 \end{array} \right.
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