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

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(B) Let the perpendicular from the right-angle vertex to the hypotenuse be $p$. Let the two segments of the hypotenuse be $x$ and $y$. In the two smal

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

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(B) Let the perpendicular from the right-angle vertex to the hypotenuse be $p$. Let the two segments of the hypotenuse be $x$ and $y$. In the two smaller right-angled triangles formed,we have $x = p 	an 	heta$ and $y = p \cot 	heta$. The hypotenuse is $x + y = p(	an 	heta + \cot 	heta) = p(\frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta}) = \frac{p}{\sin 	heta \cos 	heta} = \frac{2p}{\sin 2	heta}$. Given that the hypotenuse is $2\sqrt{2}p$,we have $\frac{2p}{\sin 2	heta} = 2\sqrt{2}p$. $\Rightarrow \sin 2	heta = \frac{1}{\sqrt{2}}$. Thus,$2	heta = \frac{\pi}{4}$ or $2	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Therefore,$	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 perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

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