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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`

A

`60^(@)`

B

`15^(@)`

C

`75^(@)`

D

`30^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A


`Delta = (1)/(2) ab = (1)/(2) p (4p) rArr ab = 4p^(2)`
Also, `a^(2) + b^(2) = c^(2) = 16 p^(2)`
`:. (a - b)^(2) = a^(2) + b^(2) - 2ab = 8p^(2)`
Also, `(a + b)^(2) = a^(2) + b^(2) + 2ab = 24 p^(2)`
`tan.(A -B)/(2) = (a -b)/(a +b) cot.(C)/(2) = (1)/(sqrt3) xx 1`
or `(A -B)/(2) = 30^(@) " or " A - B = 60^(@)`
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