Q. In a right angled triangle the hypotenuse is $ 2\sqrt{2} $ times the length of perpendicular drawn from the opposite vertex on the hypotenuse, then the other two angles are
JamiaJamia 2009
Solution:
We have, AD = p and $ BC=2\sqrt{2}p $ Clearly, $ p=a\sin \theta =b\cos \theta $
Since, $ {{a}^{2}}+{{b}^{2}}={{(2\sqrt{2}p)}^{2}} $ $ \Rightarrow $ $ {{p}^{2}}\left[ \frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta } \right]=8{{p}^{2}} $ $ \Rightarrow $ $ 1=2{{\sin }^{2}}2\theta \Rightarrow \sin 2\theta \pm \frac{1}{\sqrt{2}} $ $ \Rightarrow $ $ \sin 2\theta =\frac{1}{\sqrt{2}}\Rightarrow 2\theta =\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{8} $ So, the other angle is $ \frac{\pi }{2}-\theta =\frac{\pi }{2}-\frac{\pi }{8}=\frac{3\pi }{8} $
