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 $\sin \left(A\right)$ is equal to

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

Answer 1

The correct option is B

$\sqrt{\frac{9a^2-c^2}{8c^2}}$

Explanation for correct option:

Step- 1: Solve for the values of $b$:

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

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

We know that if $\alpha ,\beta$ are the two roots of the quadratic equation $a{x}^2+bx+c=0$ then

1. $\alpha +\beta =\frac{-b}{a}$
2. $\alpha \beta =\frac{c}{a}$

Since $b_1,b_2$ are the two roots of this equation

Thus $b_1+b_2=2c\cos \left(A\right)$ ….(i)

$b_1{b}_2=c^2-a^2$ ….(ii)

Given that $b_2=2b_1$

Thus from (i) we have

$$\begin{gathered} \Rightarrow 3b_1=2c\cos \left(A\right)\left[\because b_2=2b_1\right] \\ \Rightarrow b_1=\frac{2c\cos \left(A\right)}{3} \end{gathered}$$

Step- 2: Solve for the required value:

From (ii) we have

$$\begin{gathered} \therefore b_1{b}_2=c^2-a^2 \\ \Rightarrow 2{\left(b_1\right)}^2=c^2-a^2\left[\because b_2=2b_1\right] \\ \Rightarrow 2{\left(\frac{2c\cos \left(A\right)}{3}\right)}^2=c^2-a^2 \\ \Rightarrow 8c^2{\cos }^2\left(A\right)=9c^2-9a^2 \\ \Rightarrow 8c^2\left(1-{\sin }^2\left(A\right)\right)=9c^2-9a^2\left[\because {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right] \\ \Rightarrow -8c^2{\sin }^2\left(A\right)=c^2-9a^2 \\ \Rightarrow {\sin }^2\left(A\right)=\frac{9a^2-c^2}{8c^2} \\ \Rightarrow \sin \left(A\right)=\sqrt{\frac{9a^2-c^2}{8c^2}} \end{gathered}$$

Hence option(B) is correct.