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In a right angled triangel the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex .One of the acute angel is
c2 = h2 + (a-x)2
b2 = h2 + x2
on adding, we have 2h2 + x2 + (a-x)2 = a2 ..(1)
also a = 4h and cos B = b/a = x/b
so x = b2/a
on solving (1), we have h2 + x2 -4hx = 0
substituting above gives
b4 + a4/16 -a2.b2 =0 .. (2)
as cos B = b/a
on solving eqn. (2) we have , Cos2B = 0.933 or 0.067
Cos B = 0.966 or 0.259
B = 15 or 75 degrees
as B is acute , B is 15 degrees
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regards
Ramesh
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