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Q.

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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a

$\frac{π}{3} and \frac{π}{6}$

b

$\frac{π}{8} and \frac{3 π}{8}$

c

$\frac{π}{4} and \frac{π}{4}$

d

$\frac{π}{5} and \frac{3 π}{10}$

answer is B.

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Detailed Solution

$\begin{array}{l} P^{2} \sec^{2} θ + P^{2} {cosec}^{2} θ = (2 \sqrt{2})^{2} P^{2} \\ \Rightarrow \frac{1}{\sin^{2} {θcos}^{2} θ} = 8 \\ \Rightarrow \sin^{2} 2 θ = {(\frac{1}{\sqrt{2}})}^{2} \end{array}$
$\begin{array}{l} \Rightarrow 2 θ = π / 4 or 3 π / 4 \\ \Rightarrow θ = π / 8, 3 π / 8 \end{array}$

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