Question

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

• A. $\frac{\pi }{3}\&\frac{\pi }{6}$
• B. $\frac{\pi }{8}\&\frac{3\pi }{8}$
• C. $\frac{\pi }{4}\&\frac{\pi }{4}$
• D. $\frac{\pi }{5}\&\frac{3\pi }{10}$

Answer 1

Answer: (B)

$AC=2\sqrt{2}P$
$\frac{P}{AD}=\frac{DC}{P}=\tan \theta$
$\because AD+DC=2\sqrt{2}P$
$\Rightarrow \frac{P}{\tan \theta }+P\tan \theta =2\sqrt{2}P$
$\Rightarrow \frac{{\cos }^2\theta +{\sin }^2\theta }{2\sin \theta \cos \theta }=\sqrt{2}$ $\Rightarrow \sin 2\theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{8}$
So $\varphi =\pi -\left(\frac{\pi }{2}+\frac{\pi }{8}\right)$
$\Rightarrow \varphi =\frac{3\pi }{8}$
Aliter : On $△ABC$
$\cos \theta =\frac{AB}{AC}=\frac{AB}{2\sqrt{2}P}$
In $△ADB$
$\sin \theta =\frac{BD}{AB}=\frac{P}{AB}$
$\therefore \sin \theta \cos \theta =\frac{1}{2\sqrt{2}}$
$\Rightarrow \sin 2\theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{6}\text{ or }\frac{3\pi }{8}$