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(B) Let the two acute angles be $x$ and $y$ such that $x > y$. In a right-angled triangle,the sum of the two acute angles is $90^{\circ}$.Given: $x -

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$4$

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(B) Let the two acute angles be $x$ and $y$ such that $x > y$. In a right-angled triangle,the sum of the two acute angles is $90^{\circ}$.Given: $x - y = 60^{\circ}$ and $x + y = 90^{\circ}$.Adding these equations: $2x = 150^{\circ} \implies x = 75^{\circ}$.Then,$y = 90^{\circ} - 75^{\circ} = 15^{\circ}$.Let $BD$ be the perpendicular from the right-angled vertex $B$ to the hypotenuse $AC$. In $	riangle ABD$,$\angle ADB = 90^{\circ}$ and $\angle BAD = y = 15^{\circ}$. Thus,$	an(15^{\circ}) = \frac{BD}{AD} \implies AD = BD \cot(15^{\circ})$.In $	riangle BDC$,$\angle BDC = 90^{\circ}$ and $\angle BCD = x = 75^{\circ}$. Thus,$	an(75^{\circ}) = \frac{BD}{CD} \implies CD = BD \cot(75^{\circ})$.The length of the hypotenuse $AC = AD + CD = BD(\cot(15^{\circ}) + \cot(75^{\circ}))$.Using $\cot(15^{\circ}) = 2 + \sqrt{3}$ and $\cot(75^{\circ}) = 2 - \sqrt{3}$,we get:$AC = BD(2 + \sqrt{3} + 2 - \sqrt{3}) = 4BD$.Therefore,the ratio $\frac{AC}{BD} = 4: 1$.

In a right-angled triangle,if the difference between the two acute angles is $60^{\circ}$,then the ratio of the length of the hypotenuse to the length of the perpendicular drawn to the hypotenuse from its opposite vertex is: (in $: 1$)

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