If in a right-angled triangle the hypotenuse | vedclass

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Question

(A) Let the right-angled triangle be $\Delta AOB$ with the right angle at $O$. Let $OC$ be the perpendicular drawn from the vertex $O$ to the hypotenu

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Let the right-angled triangle be $\Delta AOB$ with the right angle at $O$. Let $OC$ be the perpendicular drawn from the vertex $O$ to the hypotenuse $AB$,with length $OC = x$.Let $\angle OBA = 	heta$. Then $\angle OAB = 90^o - 	heta$.In $\Delta OCB$,$OC = OB \sin 	heta \Rightarrow OB = \frac{x}{\sin 	heta}$.In $\Delta OCA$,$OC = OA \cos 	heta \Rightarrow OA = \frac{x}{\cos 	heta}$.In $\Delta AOB$,the hypotenuse $AB = \sqrt{OA^2 + OB^2} = \sqrt{\frac{x^2}{\cos^2 	heta} + \frac{x^2}{\sin^2 	heta}} = x \sqrt{\frac{\sin^2 	heta + \cos^2 	heta}{\sin^2 	heta \cos^2 	heta}} = \frac{x}{\sin 	heta \cos 	heta} = \frac{2x}{\sin 2	heta}$.Given that the hypotenuse is $4$ times the perpendicular,$AB = 4x$.Therefore,$\frac{2x}{\sin 2	heta} = 4x \Rightarrow \sin 2	heta = \frac{2}{4} = \frac{1}{2}$.Since $\sin 2	heta = \sin 30^o$,we have $2	heta = 30^o$,which gives $	heta = 15^o$.The other acute angle is $90^o - 15^o = 75^o$.

If in a right-angled triangle the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex,then one of its acute angles is......$^o$

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