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 (a)`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)`

Explanation

We can create a right angle traigle `ABC` with the given details.

Please refer to video to see the diagram.

Let the length of the perpendicular drawn from `B` to `AC` is `p`.

Then, `AC = 2sqrt2p`

`BC = psec theta`

`AB = pcosec theta`

Now, `AB^2+BC^2 = AC^2`

`=> p^2cosec^2theta +p^2sec^2 theta = (2sqrt2p)^2`

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