JEE Main 2024Properties of TriangleProperties of TriangleHardQuestionIn a triangle ABC,BC=7,AC=8,AB=α∈N\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}ABC,BC=7,AC=8,AB=α∈N and cosA=23\cos \mathrm{A}=\frac{2}{3}cosA=32. If 49cos(3C)+42=mn49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}49cos(3C)+42=nm, where gcd(m,n)=1\operatorname{gcd}(m, n)=1gcd(m,n)=1, then m+nm+nm+n is equal to _________.Answer: 2Hide SolutionSolutioncosA=23=α2+82−722⋅α⋅8⇒(α2+15)3=32α3α2−32α+45=0⇒α=53,9∵α∈N⇒α=9cosC=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}cosA=32=2⋅α⋅8α2+82−72⇒(α2+15)3=32α3α2−32α+45=0⇒α=35,9∵α∈N⇒α=9cosC=2⋅7⋅872+82−92=72 cos3C=4cos3C−3cosC=4×873−6749cos3C=327−42⇒49cos3C+42=327⇒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}cos3C=49cos3C⇒⇒=4cos3C−3cosC734×8−76=732−4249cos3C+42=732m+n=39