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

Question

Ramesh V · Answer

c2 = h2 + (a-x) 2 b2 = h2 + x 2 on adding, we have 2h 2 + x 2 + (a-x) 2 = a 2 ..(1) also a = 4h and cos B = b/a = x/b so x = b 2 /a on solving (1), we have h 2 + x 2 -4hx = 0 substituting above gives b 4 + a 4 /16 -a 2 .b 2 =0 .. (2) as cos B = b/a on solving eqn. (2) we have , Cos 2 B = 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

Narendar agarwal · Answer

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 · Answer

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=tan 2 x+1/tan(x) 4/2=tan 2 x+1/2tanx 2=1/sin(2x) Sin(2x)=1/2 2x=30 x=15

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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

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

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