1.

In a right angled triangle the hypotenuse is `2sqrt(2)`times the perpendicular drawn from the opposite vertex. Then the other acuteangles 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)`

Answer» 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`
`=>cosec^2 theta+sec^2theta = 8`
`=>1/sin^2theta+1/cos^2theta = 8`
`=>cos^2theta+sin^2theta = 8sin^2thetacos^2theta`
`=>1 = 8sin^2thetacos^2theta`
`=>1 = 2(2sinthetacostheta)^2`
`=>(sin2theta)^2 = 1/2`
`=>sin2theta = 1/sqrt2`
`=>2theta = pi/4`
`=>theta = pi/8`
`:. C = pi/8`
`:. A = pi/2-pi/8 = (3pi)/8`
So, the other two acute angles will be `pi/8` and `(3pi)/8.`


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