Properties of Triangles

If $\;$ $2b = 3a$ and $\tan^2 A = \dfrac{3}{5}$, prove that there are two values to the third side, one of which is double the other.

Given: $\;$ $2b = 3a$ $\implies$ $a = \dfrac{2 b}{3}$ $\;\;\; \cdots \; (1)$

and $\;$ $\tan^2 A = \dfrac{3}{5}$

i.e. $\;$ $1 + \tan^2 A = \sec^2 A = 1 + \dfrac{3}{5} = \dfrac{8}{5}$

i.e. $\;$ $\sec A = \dfrac{2\sqrt{2}}{\sqrt{5}}$ $\implies$ $\cos A = \dfrac{\sqrt{5}}{2 \sqrt{2}}$ $\;\;\; \cdots \; (2)$

In $\triangle ABC$, by cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$

i.e. $\;$ $\dfrac{\sqrt{5}}{2 \sqrt{2}} = \dfrac{b^2 + c^2 - \dfrac{4 b^2}{9}}{2 b c}$ $\;\;\;$ [by equations $(1)$ and $(2)$]

i.e. $\;$ $\dfrac{\sqrt{5}}{\sqrt{2}} b c = \dfrac{5}{9} b^2 + c^2$

i.e. $\;$ $c^2 - \left(\dfrac{\sqrt{5}}{\sqrt{2}}b\right)c + \dfrac{5}{9}b^2 = 0$

i.e. $\;$ $c = \dfrac{\dfrac{\sqrt{5}}{\sqrt{2}}b \pm \sqrt{\dfrac{5b^2}{2} - \dfrac{20 b^2}{9}}}{2}$

i.e. $\;$ $c = \dfrac{\dfrac{\sqrt{5}}{\sqrt{2}}b \pm \dfrac{\sqrt{5}}{3 \sqrt{2}}b}{2}$

i.e. $\;$ $c = \dfrac{\sqrt{5}}{2 \sqrt{2}} b \pm \dfrac{\sqrt{5}}{6 \sqrt{2}}b$

i.e. $\;$ $c = \dfrac{4 \sqrt{5}}{6 \sqrt{2}}b$ $\;$ or $\;$ $c = \dfrac{2 \sqrt{5}}{6 \sqrt{2}}b$

i.e. $\;$ $c = 2 \left(\dfrac{\sqrt{5}}{3 \sqrt{2}}b\right)$ $\;$ or $\;$ $c = \dfrac{\sqrt{5}}{3 \sqrt{2}}b$

$\therefore \;$ There are two values to the third side $c$, one of which is double the other.

Hence proved.