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QJEEM24Maa129: JEE Main 2024 Mathematics - Class 11 JEEM24Maa129 Solution: In a triangle and . If , where , then is equal to _________.

In a triangle ABC,BC=7,AC=8,AB=α∈N\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N} and cos⁡A=23\cos \mathrm{A}=\frac{2}{3} . If 49cos⁡(3C)+42=mn49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}} , where gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1 , then m+nm+n is equal to _________.

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JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Properties of Triangle Question 3 English Explanation

amp;cos⁡A=23=α2+82−722⋅α⋅8amp;⇒(α2+15)3=32αamp;3α2−32α+45=0amp;⇒α=53,9amp;∵α∈N⇒α=9amp;cos⁡C=72+82−922⋅7⋅8=27\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}

cos⁡3Camp;=4cos⁡3C−3cos⁡C=amp;4×873−6749cos⁡3Camp;=327−42⇒amp;49cos⁡3C+42=327⇒amp;m+n=39\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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