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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 -- regards Ramesh
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
--
regards
Ramesh
As shown in above figure of previous answerof same question.. Let angle B= @ then angle A= 90-@Aplly sine rule in right triangle PBC (SineB/ h) = (sineh/ /b)i.eSin@/h = sin90/b Sin@ = h/b.......eqtn 1 Now in triangle right angled triangle ABC SineA/b = sineC/aSine(90-@)/b = sin90/a Cos@/b= sin90/aThere foreCos@=b/a ... ..eqtn 2 Multiply eqntn1&2Sin@ cos@ = (h/b) * (b/a) Sin@ cos@ = h/aSin@ cos@ = h/4h a=4h given in qstnSin@ cos@ = ¼Multiply 2 both side we get2sin@cos@=2/4 Sin2@=1/2Sun2@= sin30 2@=30@=15 Ans.. RegardNarendar
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