JEE Mains · Maths · STD 11 - Trigonometrical equations
In a triangle \(\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}\) and \(\cos A=\frac{2}{3}\). If \(49 \cos (3 C)+42=\frac{m}{n}\), where \(\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\), then \(\mathrm{m}+\mathrm{n}\) is equal to ..........
- A \(29\)
- B \(39\)
- C \(40\)
- D \(31\)
Answer & Solution
Correct Answer
(B) \(39\)
Step-by-step Solution
Detailed explanation
\( \cos \mathrm{A}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}} \) \( \frac{2}{3}=\frac{8^2+\mathrm{c}^2-7^2}{2 \times 8 \times \mathrm{c}} \) \( \mathrm{C}=9 \) \( \cos \mathrm{C}=\frac{7^2+8^2-9^2}{2 \times 7 \times 8}=\frac{2}{7} \)…
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