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Grade: 12
        

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

10 years ago

Answers : (2)

Ramesh V
70 Points
							

5674-390_5850_untitled.JPG

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


10 years ago
Narendar agarwal
13 Points
							
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 
 
 
 
 
 
 
 
one year ago
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