In a right angled triangle the hypotenuse is `2sqrt(2)` times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are `pi/3a n dpi/6` (b) `pi/8a n d(3pi)/8` (c) `pi/4a n dpi/4` (d) `pi/5a n d(3pi)/(10)`

Options

A. `(pi)/(3)` and `(pi)/(6)`
B. `(pi)/(8)` and `(3pi)/(8)`
C. `(pi)/(4)` and `(pi)/(4)`
D. `(pi)/(5)` and `(3pi)/(10)`

Answer

B

Explanation

`p^(2)sec^(2)theta+p^(2)co sec^(2)theta=(2sqrt(2))^(2)p^(2)`
`rArr(1)/(sin^(2)thetacos^(2)theta)=8`
`rArr sin^(2)2theta=((1)/(sqrt(2)))^(3)`
`rArr 2theta=pi//4` or `3pi//4`
`rArr theta=pi//8,3pi//8`