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.e
Sin@/h = sin90/b 
Sin@ = h/b.......eqtn 1 
 
Now in triangle  right angled triangle ABC
 
SineA/b = sineC/a
Sine(90-@)/b = sin90/a
 
Cos@/b= sin90/a
There fore
Cos@=b/a ... ..eqtn 2
 
Multiply eqntn1&2
Sin@ cos@ = (h/b) * (b/a)
 
Sin@ cos@ = h/a
Sin@ cos@ = h/4h               a=4h  given in qstn
Sin@ cos@ = ¼
Multiply 2 both side we get
2sin@cos@=2/4
 
Sin2@=1/2
Sun2@= sin30
 
2@=30
@=15  Ans..
 
Regard
Narendar 
 
 
 
 
 
 
 
						

							
Same fig as the first answer
In triangle ABC let angle B be x
Then angle A will be 90-x
In triangle CPB 
Tan(B)=CP/PB
Tan(x)=CP/PB
PB=CPcot(x)
In triangle ACP
  tan(A)=CP/AP
Tan(90-x)=CP/AP
Cot(x)=CP/AP
AP=CPtan(x)
Now,
AB=AP+PB
4CP=CPtan(x)+CPcot(x)
4=tan(x)+cot(x)
4=tan(x)+1/tan(x)
4=tan2x+1/tan(x)
4/2=tan2x+1/2tanx
2=1/sin(2x)
Sin(2x)=1/2
2x=30
x=15