Question

The sides of a right angled triangle form a G.P. The tangent of the smallest angle is

• A. $\sqrt{\frac{\sqrt{5}+1}{2}}$
• B. $\sqrt{\frac{\sqrt{5}-1}{2}}$
• C. $\sqrt{\frac{2}{\sqrt{5}+1}}$
• D. $\sqrt{\frac{2}{\sqrt{5}-1}}$

Answer 1

The correct option is D $\sqrt{\frac{2}{\sqrt{5}-1}}$

Let, sides of right angled triangle $\Delta$ ABC are $a$, $ar$ and $a{r}^2$.Assume $r>1$.
$\therefore (a{r}^2{)}^2=a^2+a^2{r}^2$
or, $a^2{r}^4=a^2+a^2{r}^2$
or, $r^4=1+r^2$
or, $r^4-r^2-1=0$
Let, $r^2=t$, then
$t^2-t-1=0$
or, $t=\frac{1\pm \sqrt{1^2-4\times \left(1\right)\times (-1)}}{2\times 1}$
or, $t=\frac{1\pm \sqrt{5}}{2}$
or,$t=\frac{1+\sqrt{5}}{2}$ and $t=\frac{1-\sqrt{5}}{2}$
or,$r^2=\frac{1+\sqrt{5}}{2}$ and $r^2=\frac{1-\sqrt{5}}{2}$ [Not possible]
$\therefore r^2=\frac{1+\sqrt{5}}{2}$
$\Rightarrow r=\sqrt{\frac{1+\sqrt{5}}{2}}=\sqrt{\frac{2}{\sqrt{5}-1}}$