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JEE Main 2024
Properties of Triangle
Properties of Triangle
Hard

Question

In a triangle ABC,BC=7,AC=8,AB=αN\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N} and cosA=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 _________.

Answer: 2

Solution

cosA=23=α2+82722α8(α2+15)3=32α3α232α+45=0α=53,9αNα=9cosC=72+8292278=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} cos3C=4cos3C3cosC=4×8736749cos3C=3274249cos3C+42=327m+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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