If the hypotenuse of a right-angled triangle is four times the length of the perpendicular drawn from the opposite vertex to it, then the difference of the two acute angles will be `60^0` (b) `15^0` (c) `75^0` (d) `30^0`

Explanation

`/_=1/2*ab=1/2*P(4P)`

`ab=4P^2-(1)`

`a^2+b^2=c^2=16P^2`

`(a-b)^2+2ab=16P^2`

`(a-b)^2=8P^2`

`tan((A-B)/2)=(a-b)/(a+b)cos(c/2)`

`tan((A-B)/2)=(2sqrt2P)/(2sqrt2sqrt3P)=1/sqrt3`

`((A-B)/2)=pi/6=60^@`.