Question

In a $△ABC,a,c,A$ are given and $b_2=2b_1$ where $b_1,b_2$ are two values of the third side, then prove that $3a=c\sqrt{(1+8si{n}^{2}A)}$.

Answer 1

Here the quadratic for third side b is given by
$b^2-2bccosA+(c^2-a^2)=0$
$\therefore b_1+b_2=2ccosA$ ...........(1)
and $b_1{b}_2=c^2-a^2$ .......(2)
Also it is given that
$b_2=2b_1$ ...........(3)
Hence from (1) and (3),
$3b^1=2ccosA$ ........(4)
$2b_1^2=c^2-a^2$ .....(5)
Finally (4) and (5), we have
$2.\frac{4c^{2}co{s}^{2}A}{9}=c^2-a^2$
or $8c^2(1-si{n}^{2}A)=9c^2-9a^2$
or $9a^2=c^2(1+8si{n}^{2}A)$
Hence $3a=c\sqrt{1+8si{n}^{2}A}$.