In a triangle ABC, BC = 7, AC = 8, and If where then is equal to

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Solution

Q.
In a triangle ABC, BC = 7, AC = 8, $A B = α \in N$ and $\cos A = \frac{2}{3} .$ If $49 \cos (3 C) + 42 = \frac{m}{n},$ where $\gcd (m, n) = 1,$ then $m + n$ is equal to ________ . [2024]

Ans.
(39)

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{8^{2} + α^{2} - 7^{2}}{2 \cdot 8 \cdot α}$

$\Rightarrow \frac{2}{3} = \frac{α^{2} + 15}{16 α}$

$\Rightarrow 32 α = 3 α^{2} + 45$

$\Rightarrow 3 α^{2} - 32 α + 45 = 0 \Rightarrow α = \frac{5}{3}, 9$

$\Rightarrow α = 9 [∵ α \in N]$

$Now, \cos C = \frac{7^{2} + 8^{2} - 9^{2}}{2 \times 7 \times 8} \Rightarrow \cos C = \frac{2}{7}$

$Now, \cos 3 C = 4 \cos^{3} C - 3 \cos C = \frac{4 \times 8}{7^{3}} - \frac{6}{7}$

$So, 49 \cos 3 C + 42 = 49 (\frac{4 \times 8}{7^{3}} - \frac{6}{7}) + 42 = \frac{32}{7} - 42 + 42 = \frac{m}{n}$

$\Rightarrow m = 32, n = 7; m + n = 32 + 7 = 39$

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