Given b+c 11 = c+a 12 = a+b 13 for a ABC with usual notation. If A = B = C , then the…
Mathematics · JEE Main · NTA Exams — Trigonometry
Given \(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}\) for a ΔABC with usual notation. If \(\frac{\cos A}{\square}=\frac{\cos B}{\beta}=\frac{\cos C}{\gamma}\), then the ordered triad (α, β, γ) has a value
- (19, 7, 25)
- (7, 19, 25)
- (5, 12, 13)
- (3, 4, 5)
Answer
(B) (7, 19, 25)
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