question_answer

    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

A)  \[\frac{\pi }{3},\frac{\pi }{6}\]                   

B)  \[\frac{\pi }{4},\frac{\pi }{4}\]   

C)  \[\frac{\pi }{8},\frac{3\pi }{8}\]                

D)  \[\frac{\pi }{12},\frac{5\pi }{12}\]

Correct Answer: C

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 \[\frac{\pi }{2}-\theta =\frac{\pi }{2}-\frac{\pi }{8}=\frac{3\pi }{8}\]