In $△ A B C, ∠ B = 90 °$ and perpendicular from $B$ on $A C$ intersects it at $D .$ If $A C = 4 B D,$ then the smallest angle of $△ A B C$ is

\frac{π}{12}

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\frac{π}{8}

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\frac{π}{6}

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\frac{π}{5}

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Solution

The correct option is A $\frac{π}{12}$
Let $B D = x, \Rightarrow A C = 4 x$

In $△ B C D$
$sin θ = \frac{x}{a}$
In $△ A B C$
$cos θ = \frac{a}{4 x}$
$\Rightarrow sin θ cos θ = \frac{1}{4} \Rightarrow sin 2 θ = \frac{1}{2} \Rightarrow 2 θ = \frac{π}{6}$
$∴ θ = \frac{π}{12}$

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