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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
Trigonometric Functions
Solution:
$ 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 $\phi=\pi-\left(\frac{\pi}{2}+\frac{\pi}{8}\right) $
$ \Rightarrow \phi=\frac{3 \pi}{8}$
Aliter : On $\triangle ABC$
$\cos \theta=\frac{A B}{A C}=\frac{A B}{2 \sqrt{2} P}$
In $\triangle A D B$
$ \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}} $
$\Rightarrow \theta=\frac{\pi}{6} \text { or } \frac{3 \pi}{8}$