50 Follower Special

It is a right triangle so $q^2 + t^2 = g^2,$ also by Vieta's formula $\frac{-g}{q} = -1.25 \Rightarrow g = \frac{5q}{4}.$ Lastly $t = \sqrt{g^2 - q^2} = \sqrt{\frac{9q^2}{16}} = \frac{3q}{4}.$ So the ratio $q:g:t$ is $4:5:3$ and $\lfloor x \rfloor = \lfloor \sin^{-1} \frac{3}{5}\rfloor = 36^\circ.$ The digit product is $3\times 6 = 18.$

given $\frac{g}{q}=1.25$ ....... $(1)$

using sine rule

$\frac{q}{sinQ}=\frac{g}{sinG}=\frac{t}{sinT}$

from this $\frac{g}{sin90^{o}}=\frac{q}{sinQ}$

$sinQ=\frac{1}{1.25}$

$\angle Q=53.13^{o}$

$\angle T=180^{o}-(90^{o}+53.13^{o})$

$\angle QTG=36.87^{o}$

$[36.87]=36$

hence product of digits $3\times 6=18$

Using Vieta's formula:

$\large -1.25=-\frac{g}{q} \Rightarrow 1.25=\frac{g}{q}$ .

From the right triangle we can show that:

$\large cos(x)=\frac{q}{g}=\frac{1}{1.25}=\frac{4}{5}$ .

$\large \therefore \left\lfloor x \right\rfloor =\left\lfloor arccos(\frac { 4 }{ 5 } ) \right\rfloor = 36°$ .

Therefore the digit product is $\large \color{#20A900}{\boxed {18}}$ .

g/q=1.25 then cos x =q/g=1/1.25=.8 then x=36.86 (approximately). Then floor function of x is 36 then the product of their digit is 18