Question

In a right angled triangle, the hypotenuse is $2\sqrt{2}$ times the length of the 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)

$\frac{\pi }{8},\frac{3\pi }{8}$

In $△ABC$
Let $\angle ACB=\theta$ and $BD=x\Rightarrow AC=2\sqrt{2}x$
$\cos \theta =\frac{BC}{AC}=\frac{BC}{2\sqrt{2}x}...\left(1\right)$
And in $△BCD$
$\sin \theta =\frac{BD}{BC}=\frac{x}{BC}...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$2\sin \theta \cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{8}$

$\angle BAC=\frac{\pi }{2}-\frac{\pi }{8}=\frac{3\pi }{8}$