JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 22)
In a triangle $$\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}$$ and $$\cos \mathrm{A}=\frac{2}{3}$$. If $$49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}$$, where $$\operatorname{gcd}(m, n)=1$$, then $$m+n$$ is equal to _________.
Answer
39
Explanation
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$$\begin{aligned} & \cos A=\frac{2}{3}=\frac{\alpha^2+8^2-7^2}{2 \cdot \alpha \cdot 8} \\ & \Rightarrow\left(\alpha^2+15\right) 3=32 \alpha \\ & 3 \alpha^2-32 \alpha+45=0 \\ & \Rightarrow \alpha=\frac{5}{3}, 9 \\ & \because \alpha \in N \Rightarrow \alpha=9 \\ & \cos C=\frac{7^2+8^2-9^2}{2 \cdot 7 \cdot 8}=\frac{2}{7} \end{aligned}$$
$$ \begin{aligned} \cos 3 C & =4 \cos ^3 C-3 \cos C \\ = & \frac{4 \times 8}{7^3}-\frac{6}{7} \\ 49 \cos 3 C & =\frac{32}{7}-42 \\ \Rightarrow \quad & 49 \cos 3 C+42=\frac{32}{7} \\ \Rightarrow \quad & m+n =39 \end{aligned} $$
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