Question

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. $\frac{\pi }{12},\frac{5\pi }{12}$

Answer 1

Answer: (C)

We have, AD = p and $BC=2\sqrt{2}p$ Clearly, $p=a\sin \theta =b\cos \theta$
Since, $a^2+b^2={(2\sqrt{2}p)}^2$ $\Rightarrow$ $p^2\left[\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }\right]=8p^2$ $\Rightarrow$ $1=2{\sin }^{2}2\theta \Rightarrow \sin 2\theta \pm \frac{1}{\sqrt{2}}$ $\Rightarrow$ $\sin 2\theta =\frac{1}{\sqrt{2}}\Rightarrow 2\theta =\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{8}$ So, the other angle is $\frac{\pi }{2}-\theta =\frac{\pi }{2}-\frac{\pi }{8}=\frac{3\pi }{8}$