Christmas Math Problem

Since the hypotenuse has length $21$ and one side $n$ , therefore, the length of other side is $\sqrt{441-n^2}$ by pythagoras theorem.

To solve for $n$ , we must solve the given quadratic equation which is,

$\begin{aligned} & 9n^2 - 279n +2162.25 = 0 \\ \implies & n^2 - 31n + \dfrac{961}{4} = 0 \\ \implies & {\left(n - \dfrac{31}{2} \right)}^2 = 0 \\ \implies & n = \dfrac{31}{2} \end{aligned}$

Therefore, the length of other side will be,

$\sqrt{441 - {\left(\dfrac{31}{2}\right)}^2} = \dfrac{\sqrt{803}}{2}$

Using cosine rule, we have,

$\cos A = \dfrac{ 21^2 + {\left(\frac{\sqrt{803}}{2}\right)}^2 - {\left(\frac{31}{2}\right)}^2 }{ 2 \times 21 \times \frac{\sqrt{803}}{2} } = \dfrac{\sqrt{803}}{42}$

$\cos B = \dfrac{ 21^2 - {\left(\frac{\sqrt{803}}{2}\right)}^2 + {\left(\frac{31}{2}\right)}^2 }{ 2 \times 21 \times \frac{31}{2} } = \dfrac{31}{42}$

Therefore,

$A = \cos^{-1} \left(\dfrac{\sqrt{803}}{42}\right) \approx 0.83 \ \text{rad} \\ B = \cos^{-1} \left(\dfrac{31}{42}\right) \approx 0.74 \ \text{rad}$

And, $\dfrac{A}{B} = \dfrac{0.83}{0.74} \approx \boxed{1.12}$