Question

Let $1<m<3$ In a triangle ABC if $2b=(m+1)a$ and $\cos A=\frac{1}{2}\sqrt{\frac{(m-1)(m+3)}{m}}$.
Prove that there are two values of the third side one of which is $m$ times the other.

Answer 1

Given,

$2b=(m+1)a$ …… (1)

And $4m{\cos }^{2}a=(m-1)(m+3)$ ……. (2)

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

From equation (1) $\cos A=\frac{{\left(\frac{(m+1)a}{2}\right)}^2+c^2-a^2}{(m+1)ac}$

$(m+1)ac\cos A={(m+1)}^2.\frac{a^2}{4}+c^2-a^2$

$m\times ac\cos A+a.c.\cos A=\left(\frac{{(m+1)}^2}{4}\right)a^2-a^2+c^2$

$m\times accosA+ac\cos A=({(m+1)}^2-2^2)\frac{a^2}{4}+c^2$

$m\times accosA+ac\cos A=\left(\right(m-1\left)\right(m+3\left)\right)\frac{a^2}{4}+c^2$

From equation $\left(2\right)$

$m\times ac\times cosA+ac\cos A=(4.m.{\cos }^{2}A)\frac{a^2}{4}+c^2$

$m.a.c.cosA+a.c.\cos A=(m.{\cos }^{2}A)a^2+c^2$

$m.a.c.cosA+a.c.\cos A-a^2.m.{\cos }^{2}A-c^2=0$

$m.a.c.cosA-a^{2}m{\cos }^{2}A+ac.\cos A-c^2=0$

$a.m.\cos A(c-a\cos A)-c(c-a\cos A)=0$

$(c-a\cos A)(am.\cos A-c)=0$

When,

$c-a\cos A=0$

$c=a\cos A$

When,

$am.\cos A-c=0$

$c=m.a\cos A$

Hence proved there are two values of third side one of which is $m$ times of the other.