In a right-angled triangle,the hypotenuse is | vedclass

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(C) Let the right-angled triangle have sides $a$ and $b$ and hypotenuse $c$. Let $p$ be the length of the perpendicular from the right-angle vertex to

Vedclass · Accepted Answer

$\frac{\pi}{8}, \frac{3\pi}{8}$

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(C) Let the right-angled triangle have sides $a$ and $b$ and hypotenuse $c$. Let $p$ be the length of the perpendicular from the right-angle vertex to the hypotenuse.We know that $p = \frac{ab}{c}$.Given $c = 2\sqrt{2}p$,we have $c = 2\sqrt{2} \frac{ab}{c}$,so $c^2 = 2\sqrt{2}ab$.Since $c^2 = a^2 + b^2$,we have $a^2 + b^2 = 2\sqrt{2}ab$.Dividing by $ab$,we get $\frac{a}{b} + \frac{b}{a} = 2\sqrt{2}$.Let $	heta$ be one of the acute angles,so $	an 	heta = \frac{b}{a}$.Then $\cot 	heta + 	an 	heta = 2\sqrt{2}$.$\frac{1}{	an 	heta} + 	an 	heta = 2\sqrt{2} \Rightarrow \frac{1 + 	an^2 	heta}{	an 	heta} = 2\sqrt{2}$.$\frac{\sec^2 	heta}{	an 	heta} = 2\sqrt{2}$ $\Rightarrow \frac{1}{\sin 	heta \cos 	heta} = 2\sqrt{2}$ $\Rightarrow \frac{2}{\sin 2	heta} = 2\sqrt{2}$.$\sin 2	heta = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$.Thus,$2	heta = \frac{\pi}{4}$ or $2	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.So,$	heta = \frac{\pi}{8}$ or $	heta = \frac{3\pi}{8}$.The two acute angles are $\frac{\pi}{8}$ and $\frac{3\pi}{8}$.

In a right-angled triangle,the hypotenuse is $2\sqrt{2}$ times the length of the perpendicular drawn from the opposite vertex to the hypotenuse. Then,the other two angles are:

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