Question

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

Then $\sin A$ is equal to?

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

Answer 1

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

We have, $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow b^2-2bc\cos A+(c^2-a^2)=0$

It is given that $b_1$ and $b_2$ are roots of the above equation.

Therefore $b_1+b_2=2c\cos A$ and $b_1{b}_2=c^2-a^2$

$\Rightarrow 3b_1=2c\cos A$ and $2b_1^2=c^2-a^2$ $($ Since $b_2=2b_1)$

$\Rightarrow 2{\left(\frac{2c}{3}\cos A\right)}^2=c^2-a^2$

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

$\Rightarrow \sin A=\sqrt{\frac{9a^2-c^2}{8c^2}}$