Question

The three sides of a right-angled triangle are in G.P (geometric progression). If the two acute angles be $\alpha$ and $\beta$, then tan $\alpha$ and tan $\beta$ are

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

Answer 1

Answer: (B)

$\sqrt{\frac{\sqrt{5}+1}{2}}\,and\, \sqrt{\frac{\sqrt{5}-1}{2}}$

Answer 2

$\left(\frac{a}{r}\right)^{2} +a^{2} = a^{2}r^{2} \,\left(r > 1\right), \,a \ne 0\,\Rightarrow\,r^{4}-r^{2}-1 = 0\,\Rightarrow r^{2} = \frac{1\pm\sqrt{5}}{2}\,\Rightarrow r = \pm\sqrt{\frac{\sqrt{5}+1}{2}}$
$\Rightarrow r = \sqrt{\frac{\sqrt{5}+1}{2}} \left(r > 1\right), \frac{1}{r} = \sqrt{\frac{\sqrt{5}-1}{2}}\left(\because\,\alpha+\beta = 90^{\circ} \Rightarrow tan\,\alpha = cot\,\beta = \frac{1}{tan\,\beta} \right)$