In a right angled  Δle, the hypotenuse is  22 times the length of perpendicular drawn from the opposite vertex to the hypotenuse, then the other two angles are

# In a right angled  ${\Delta }^{le}$, the hypotenuse is  $2\sqrt{2}$ times the length of perpendicular drawn from the opposite vertex to the hypotenuse, then the other two angles are

1. A

$\frac{\pi }{3},\frac{\pi }{6}$

2. B

$\frac{\pi }{4},\frac{\pi }{4}$

3. C

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

4. D

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

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### Solution:

$\mathrm{tan}A=\frac{x}{AM}⇒AM=\frac{x}{\mathrm{tan}A}$

Similarly $CM=\frac{x}{\mathrm{tan}C}$

$AM+CM=2\sqrt{2}x$

$2\sqrt{2}x=\frac{x}{\mathrm{tan}A}+\frac{x}{\mathrm{tan}C}$

$2\sqrt{2}=\frac{1}{\mathrm{tan}A}+\frac{1}{\mathrm{tan}C}$

$\mathrm{tan}A+\mathrm{tan}C=2\sqrt{2}\mathrm{tan}A\mathrm{tan}C$

Also  $A+C=\frac{\pi }{2}$

$\frac{\mathrm{tan}A+\mathrm{tan}C}{1-\mathrm{tan}A\mathrm{tan}C}=\infty$

$\mathrm{tan}A\mathrm{tan}C=1$

$\mathrm{tan}A+\mathrm{tan}C=2\sqrt{2}$

$\mathrm{tan}A\mathrm{tan}C=1$

By solving these two we get  $\mathrm{tan}A=\sqrt{2}+1$

$\mathrm{tan}C=\sqrt{2}-1$

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