Q46 / 120 · VITEEE 2008 · Mathematics
In a triangle $\mathrm{ABC}$, the sides $\mathrm{b}$ and $\mathrm{c}$ are the roots of the equation $x^{2}-61 x+820=0$ and $A=\tan ^{-1}\left(\frac{4}{3}\right)$, then $a^{2}$ is equal to
- A1098
- B1096
- Correct answerC1097
- D1095
Solution
Since b and $\mathrm{C}$ are the roots of
$\begin{array}{l}
\mathrm{x}^{2}-61 \mathrm{x}+820=0, \text { so } \\
\mathrm{b}+\mathrm{c}=61, \mathrm{bc}=820 \\
\mathrm{~A}=\tan ^{-1}\left(\frac{4}{3}\right) \Rightarrow \tan \mathrm{A}=\frac{4}{3} \Rightarrow \cos \mathrm{A}=\frac{3}{5}
\end{array}$
Now, using the formula,
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$\begin{aligned} \Rightarrow \mathrm{a}^{2} &=\mathrm{b}^{2}+\mathrm{c}^{2}-2 \mathrm{bc} \cos \mathrm{A} \\ &=(\mathrm{b}+\mathrm{c})^{2}-2 \mathrm{bc}-2 \mathrm{bc} \cos \mathrm{A} \\ &=(\mathrm{b}+\mathrm{c})^{2}-2 \mathrm{bc}(1+\cos \mathrm{A}) \\ &=(61)^{2}-2 \times 820\left(1+\frac{3}{5}\right) \\ &=3721-2624=1097 \end{aligned}$
Hence, the correct answer is option: 'C'.