Question

The three sides of a right-angled triangle are in G .P.. The tangents of the two acute angles may be-

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

Answer 1

Answer: (B), (D)

The correct options are
B $\sqrt{\frac{(\sqrt{5}-1)}{2}}$
D $\sqrt{\frac{(\sqrt{5}+1)}{2}}$

Let the sides be $a,ar,a{r}_2$ and two acute angles be \alpha, \beta . If r > 1, then
$(a{r}^2{)}^2=(a{)}^2+(ar{)}^2$ (since in this case $(ar{)}^2$ will be the hypotenuse i.e., the largest side)
$\Rightarrow r^4=1+r^2\Rightarrow r^2=\frac{1+\sqrt{5}}{2},(r^2=\frac{1-\sqrt{5}}{2})$ is not possible.
If 0 < r < 1 then a is the largest side
$\Rightarrow r=tan\left(\alpha \right)=\sqrt{\frac{1+\sqrt{5}}{2}}$
$\Rightarrow \frac{1}{r}=tan\left(\beta \right)=\sqrt{\frac{\sqrt{5}-1}{2}}$
Hence, options 'B' and 'D' are correct.