# 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

Ramesh V
70 Points
13 years ago

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

Narendar agarwal
13 Points
4 years ago
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

Shruti
13 Points
3 years ago
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