Mathematics · 2019
JEE Main · 11 January 2019, Shift 2 · Q88
Given (b + c)/11=(c + a)/12=(a + b)/13 for a ABC with usual notation. If (cos A)/(α)=(cos B)/(β)=(cos C)/(γ), then the ordered triad (α, β, γ) has a…
Given $\displaystyle \frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}$ for a $\displaystyle \triangle \mathrm{ABC}$ with
usual notation. If $\displaystyle \frac{\cos \mathrm{A}}{\alpha}=\frac{\cos \mathrm{B}}{\beta}=\frac{\cos \mathrm{C}}{\gamma}$,
then the ordered triad $\displaystyle (\alpha, \beta, \gamma)$ has a value:
Official answer
(3)
($\displaystyle 7$, $\displaystyle 19$, $\displaystyle 25$)
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.