The sides of a right triangle form a G.D. The tangent of the smallest angle is 1.1 (A) \( \sqrt{\frac{\sqrt{5}+1}{2}} \) (B) \( \sqrt{\frac{\sqrt{5}-1}{2}} \) (C) \( \sqrt{\frac{2}{\sqrt{5}+1}} \) (D) \( \sqrt{\frac{2}{\sqrt{5}-1}} \)

Question

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

Answer 1

Correct option is (D) \(\sqrt {\frac {2}{\sqrt {5}-1}}\)

Let, sides of right angled triangle \(\Delta \mathrm{ABC}\) are \(\mathrm{a},\) ar and \(\mathrm{ar}^2\). Assume \(\mathrm{r}>1\).

\(\therefore\left(a r^2\right)^2=a^2+a^2 r^2\)

or, \(a^2 r^4=a^2+a^2 r^2\)

or, \(\mathrm{r}^4=1+\mathrm{r}^2\)

or, \(\mathrm{r}^4-\mathrm{r}^2-1=0\)

Let, \(\mathrm{r}^2=\mathrm{t},\) then

\(\mathrm{t}^2-\mathrm{t}-1=\mathrm{O}\)

or, \(t=\frac{1 \pm \sqrt{1^2-4 \times(1) \times(-1)}}{2 \times 1}\)

or, \(t=\frac{1 \pm \sqrt{5}}{2}\)

or, \(\mathrm{t}=\frac{1+\sqrt{5}}{2}\) and \(\mathrm{t}=\frac{1-\sqrt{5}}{2}\)

or, \(\mathrm{r}^2=\frac{1+\sqrt{5}}{2}\) and \(\mathrm{r}^2=\frac{1-\sqrt{5}}{2}\) [Not possible]

\(\therefore \mathrm{r}^2=\frac{1+\sqrt{5}}{\underline{2}}\)

\(\Rightarrow r=\sqrt{\frac{1+\sqrt{5}}{2}}=\sqrt{\frac{2}{\sqrt{5}-1}}\)