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

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(B) Let the perpendicular from the right-angled vertex to the hypotenuse be $p$.From the geometry of the triangle,the two segments of the hypotenuse a

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

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(B) Let the perpendicular from the right-angled vertex to the hypotenuse be $p$.From the geometry of the triangle,the two segments of the hypotenuse are $p 	an 	heta$ and $p \cot 	heta$,where $	heta$ is one of the acute angles.The hypotenuse $h = p 	an 	heta + p \cot 	heta = p(	an 	heta + \cot 	heta) = p \left( \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} ight) = p \left( \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} ight) = \frac{p}{\sin 	heta \cos 	heta} = \frac{2p}{\sin 2	heta}$.Given that $h = 2\sqrt{2} p$,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 $2	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.$\Rightarrow 	heta = \frac{\pi}{8}$ or $	heta = \frac{3\pi}{8}$.Since the sum of acute angles in a right-angled triangle is $\frac{\pi}{2}$,the 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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