Question

In a $△ABC$ the values of $a,c,A$ are given and $b_1,b_2$ are two values of the third side $b$ such that $b_2=2b_1$. Then $\sin A=?$

• A. $\sqrt{\frac{9a^2-c^2}{8a^2}}$
• B. $\sqrt{\frac{9a^2-c^2}{8c^2}}$
• C. $\sqrt{\frac{9a^2+c^2}{8a^2}}$
• D. None of these

Answer 1

The correct option is B $\sqrt{\frac{9a^2-c^2}{8c^2}}$

We know, $\cos A=\frac{b^2+c^2-a^2}{2bc}\Rightarrow b^2-\left(2c\cos A\right)b+(c^2-a^2)=0$

Its roots are $b_1,b_2$

$\Rightarrow b_1+b_2=2\cos A;b_1{b}_2=(c^2-a^2)$

Also given $b_2=2b_1\Rightarrow b_1=\frac{2c\cos A}{3};2{b_1}^2=c^2-a^2$

$\Rightarrow 2\left(\frac{4c^2{\cos }^{2}A}{9}\right)=c^2-a^2$

$\Rightarrow {\cos }^{2}A=\frac{9}{8}\left(\frac{c^2-a^2}{c^2}\right)$

$\Rightarrow \sin A=\sqrt{1-\frac{9}{8}\left(\frac{c^2-a^2}{c^2}\right)}=\sqrt{\frac{9a^2-c^2}{8c^2}}$