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

\frac{π}{3}

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\frac{π}{8}

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\frac{3 π}{8}

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\frac{π}{6}

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Solution

The correct options are
B $\frac{π}{8}$
C $\frac{3 π}{8}$
Given: $A C = 2 \sqrt{2} B D$ ... $(1)$
From the Figure:
In right angled triangle $B A C$
$B D ⊥ A C$
Let, $∠ A B D = θ$ and $C B D = ϕ$
In right angled triangle $A D B$ :
From Sine Rule
$\frac{B D}{sin A} = \frac{A D}{sin θ}$ ... $(2)$
In right angled triangle $C D B$ :
From Sine Rule
$\frac{B D}{sin C} = \frac{C D}{sin ϕ}$ ... $(3)$
From $(2)$ and $(3)$
$A D + C D = B D (\frac{sin θ}{sin A} + \frac{sin ϕ}{sin C})$
$A C = \frac{A C}{2 \sqrt{2}} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{sin (\frac{π}{2} - A)}{sin A} + \frac{sin (\frac{π}{2} - \frac{π}{2} + A)}{sin (\frac{π}{2} - A)} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$ ....{ from $(1)$ }
$\Rightarrow \sqrt{2} = \frac{1}{2} (\frac{cos A}{sin A} + \frac{sin A}{cos A}) \Rightarrow 2 sin A cos A = \frac{1}{\sqrt{2}} \Rightarrow sin 2 A = \frac{1}{\sqrt{2}} \Rightarrow A = \frac{π}{8} \Rightarrow B = π - \frac{π}{2} - \frac{π}{8} = \frac{3 π}{8}$
Ans: $B, C$

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