Question

In $\Delta ABC$ $a,c$ and $A$ are given and $b_1,b_2$ are two values of the third side $b$ such that $b_2=2b_1$ Then prove that $\sin A=\sqrt{\frac{9a^2-c^2}{8c^2}}$

Answer 1

Given:

In $\Delta ABC$,

$a,c$ and $c$ are three sides.

$b_1,b_2$ are two values of the third side $b$.

$b_2=2b_1$

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

Proof:

We know that,

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

(or) $b^2-2bc\cos A+(c^2-a^2)=0$
It is given that $b_1,b_2$ are the roots of this equation.
$\Rightarrow b_1+b_2=2c\cos A,$ and $b_1{b}_2=c^2-a^2$
$3b_1=2c\cos A$ and

$2b_1^2=c^2-a^2(\because b_2=2b_{1}given)$
$2{\left(\frac{2c}{3}\cos A\right)}^2=c^2-a^2$

$\Rightarrow 4c\cos A=3(c^2-a^{2)}$

$\Rightarrow 8c^2{\cos }^{2}A=9c^2-9a^2$

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

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

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

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

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

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

Hence, proved.