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 acute angles of the triangle.

• A. $\frac{\pi }{3},\frac{\pi }{6}$
• B. $\frac{\pi }{4},\frac{\pi }{4}$
• C. $\frac{\pi }{8},\frac{3\pi }{8}$
• D. $\frac{\pi }{12},\frac{\pi }{12}$

Answer 1

The correct option is C $\frac{\pi }{8},\frac{3\pi }{8}$

Let $PQ=a$ and $QR=b$

then, $QS\perp PR$ & $QS=x$

therefore, $PR=2\sqrt{2}x$

$ar.△PQR=\frac{1}{2}\times a\times b=\frac{1}{2}\times 2\sqrt{2}x\times x$

$\Rightarrow ab=2\sqrt{2}{x}^2\Rightarrow b=\frac{2\sqrt{2}{x}^2}{a}\to 1$

Using Pythagoras theorem

${PR}^2={PQ}^2+{QR}^2\Rightarrow {\left(2\sqrt{2}x\right)}^2=a^2+b^2$

From $1$, put the value of 'b'

$a^2+{\left(\frac{2\sqrt{2}{x}^2}{a}\right)}^2=8x^2$

$a^4-8x^2{a}^2+8x^4=0$

$\Rightarrow a^2=\frac{8x^2\pm \sqrt{64x^4-32x^4}}{2}=4x^2+2\sqrt{2}{x}^2$

$\Rightarrow a=\sqrt{4x^2\pm 2\sqrt{2}{x}^2}=x\sqrt{4\pm 2\sqrt{2}}$

$\therefore b^2=8x^2-a^2=4x^2-2\sqrt{2}{x}^2$

$4x^2+2\sqrt{2}{x}^2$

From $(+)$ value of $a,b=\sqrt{4x^2-2\sqrt{2}{x}^2}=x\sqrt{4-2\sqrt{2}}$

and for $(-)$ value of $a,b=\sqrt{4x^2+2\sqrt{2}{x}^2}=x\sqrt{4+2\sqrt{2}}$

$△PQR$,

$\tan R=\frac{a}{b}=\frac{x\sqrt{4+2\sqrt{2}}}{x\sqrt{4-2\sqrt{2}}}=\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}$

$\tan R=\tan 22.5°\Rightarrow R=\frac{\pi }{8}$

Second value$=\frac{\pi }{2}-\frac{\pi }{8}=\frac{3\pi }{8}$