Q54 / 90 · JEE Main 2019 (11 Jan Shift 2) · Mathematics
Given $\frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}$ for a $\Delta \mathrm{ABC}$ with usual
notation. If $\frac{\cos \mathrm{A}}{\alpha}=\frac{\cos \mathrm{B}}{\beta}=\frac{\cos \mathrm{C}}{\gamma},$ then the ordered triad$(\alpha, \beta, \gamma)$ has a value
- Correct answerA(7,19,25)
- B(3,4,5)
- C(5,12,13)
- D(19,7,25)
Solution
Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$ (Say).
$\therefore \quad b+c=11 k, c+a=12 k, a+b=13 k$
$\therefore \quad a+b+c=18 k$
$\therefore \quad a=7 k, b=6 k$ and $c=5 k$
$\therefore \quad \cos A=\frac{36 k^{2}+25 k^{2}-49 k^{2}}{2.30 k^{2}}=\frac{1}{5}$
and $\cos B=\frac{49 k^{2}+25 k^{2}-36 k^{2}}{2.35 k^{2}}=\frac{19}{35}$
and $\cos C=\frac{49 k^{2}+36 k^{2}-25 k^{2}}{2.42 k^{2}}=\frac{5}{7}$
$\therefore \quad \cos A: \cos B: \cos C=7: 19: 25$
$\therefore \quad \frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$
Hence, required ordered triplet is (7,19,25)
Hence, the correct answer is option: 'A'.