The three sides of a right-angled triangle ar | vedclass

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(B) Let the sides of the right-angled triangle be $a, ar, ar^2$ where $ar^2$ is the hypotenuse.By Pythagoras theorem, $(ar^2)^2 = a^2 + (ar)^2$.Dividi

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

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(B) Let the sides of the right-angled triangle be $a, ar, ar^2$ where $ar^2$ is the hypotenuse.By Pythagoras theorem, $(ar^2)^2 = a^2 + (ar)^2$.Dividing by $a^2$ (assuming $a 
eq 0$), we get $r^4 = 1 + r^2$, which implies $r^4 - r^2 - 1 = 0$.Solving for $r^2$ using the quadratic formula, $r^2 = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 + \sqrt{5}}{2}$ (since $r^2 > 0$).Thus, $r = \sqrt{\frac{\sqrt{5}+1}{2}}$.In the triangle, $	an \alpha = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{ar}{a} = r = \sqrt{\frac{\sqrt{5}+1}{2}}$.Similarly, $	an \beta = \frac{a}{ar} = \frac{1}{r} = \frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}} = \sqrt{\frac{2}{\sqrt{5}+1}} = \sqrt{\frac{2(\sqrt{5}-1)}{5-1}} = \sqrt{\frac{\sqrt{5}-1}{2}}$.Therefore, the values are $\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$.

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

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