Question

$\text{In a triangle ABC, the sides b and c are the roots of the equation}{x}^2-61x+820=0\text{and}A=ta{n}^{-1}\frac{4}{3}\text{then}{a}^2\text{is equal to}$

• A. 1098
• B. 1096
• C. 1097
• D. 1095

Answer 1

The correct option is C

1097

$x^2-61x+820=0\Rightarrow x^2-41x-20x+820=0\Rightarrow (x-41)(x-20)=0\text{Let}b=41andc=20Also,A=ta{n}^{-1}\left(\frac{4}{3}\right)\Rightarrow tanA=\frac{4}{3}\therefore cosA=\frac{3}{5}\therefore \text{By cosine formula},a^2=b^2+c^2-2bccosA={41}^2+{20}^2-2\times 41\times 20\times \frac{3}{5}=2081-984=1097$