A geometry problem by Parth Sankhe

Let the side lengths opposite angles $A$ , $B$ and, $C$ be $a$ , $b$ , and $c$ respectively. Then $AC = b = 5$ , $AB=c=4$ and we need to find $BC=a$ .

By cosine rule : $a^2 = b^2 + c^2 - 2bc \cos A = 5^2+4^2 - 2\cdot 5\cdot 4 \cos A = 41-40\cos A \quad ...(1)$

By sine rule : $\dfrac {\sin A}a = \dfrac {\sin B}5 = \dfrac {\sin C}4 \quad ...(2)$

Consider the following identity:

$\begin{aligned} \color{#3D99F6}\cos (B-C) \color{#D61F06} - \cos (B+C) & = 2 \sin B \sin C & \small \color{#3D99F6} \text{Given that }\cos (B-C) = \frac {31}{32} \\ \color{#3D99F6}\frac {31}{32} \color{#D61F06}+ \cos A & = 2 \cdot 5 \cdot 4 \cdot \color{#20A900} \frac {\sin B}5 \cdot \frac {\sin C}4 & \small \color{#D61F06} \text{Note that }\cos A = \cos (180^\circ - B-C) = - \cos (B+C) \\ \frac {31}{32} + \cos A & = 40 \cdot \color{#20A900} \frac {\sin^2 A}{a^2} & \small \color{#20A900} \text{From }(2): \frac {\sin A}a = \frac {\sin B}5 = \frac {\sin C}4 \\ \frac {31}{32} + \cos A & = 40 \cdot \color{#20A900} \frac {1-\cos^2 A}{41-40\cos A} & \small \color{#20A900} \text{From }(1): a^2 = 41-40\cos A \\ \frac {31}{32} + \cos A & = \frac {1-\cos^2 A}{\frac {41}{40}-\cos A} \\ \left(\frac {31}{32} + \cos A\right) \left(\frac {41}{40}-\cos A \right) & = 1 - \cos^2 A \\ \frac {1271}{1280} + \frac {72}{1280}\cos A - \cos^2 A & = 1 - \cos^2 A \\ \frac {72}{1280}\cos A & = 1 - \frac {1271}{1280} \\ \cos A & = \frac 9{72} = \frac 18 \end{aligned}$

From $(1): \ a^2 = 41 - 40 \times \dfrac 18 = 36$ , $\implies a = \boxed 6$ .

Hint :- Use Napier's Analogies...............!!!!