24when A+B+C=pi
=> tanA+tanB+tanC=tanA.tanB.tanC
tanA,tanB,tanc are the roots of the eqn. x3-k2x2-px+2k+1=0,then triangle abc will be a triangle of what nature?explain the ans?
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106yes soumya.
ur correct
i should have written
tanA+tan+BtanC = 1. This implies that atleast one in tanA, tanB, tanC is negative but since tanAtanBtanC > 0 so two out of tanA, tanB, tanC is negative which is not possible for any triangle..
1That is what I am trying to say , in acute angled triangles , if tan a , tan b , tan c are all positive , then tan a tan b tan c > 3 . But here we get that tan a tan b tan c > 0 but < 3 . So no triangle is possible.
24@akari tanA +tanB +tanC >0
it wont be tanA +tanB +tanC≥0
btw nishnat sir plzz reply....the other day u said my soln was rite...
1The wrong statement is ---
If a + b + c =pi , then tan a tan b tan c >0
take for example the angles 15 , 45 , and 120 . What do you get ? A negative number !!!!
Also in an acute angled triangle , tan a tan b tan c > 3 , not > 0 , so you can't say that
tan a + tan b + tan c = 1 , which you get if you solve k^2 = -2k -1
1Ashis has written ---
tan a + tan b + tan c = 1 = tan a tan b tan c
tan a tan b tan c > 3 ( if tan a , tan b , tan c > 0 )
So one and only one of tan a , tan b , tan c is negative.
But that would imply that tan a tan b tan c < 0 , which is not correct
here as it is equal to 1 .
So either two of them are negative , which is not possible for a
triangle , or all of them are positive , which is again not possible as then
tan a tan b tan c > 1
So no such triangle is possible.
The reason that I am quite certain of my soln. , is that this ques. has
come up in one of Edudigm's quadratic practice papers , and the
answer given is no such triangle is possible .
24proof
tan(A+B+C)=S1-S31-S2
A+B+C=pi
=>tan(A+B+C)=0
=>S1=S3
=> tanA+tanB+tanC=tanA.tanB.tanC
24tanA+tanB+tanC>0
since A+B+C=pi
=> tanA.tanB.tanC >0
SO 2 cases come
a) Tow factors negative and one factor positive
b) All three factors postive
but tan takes negative value when pi>angle >pi/2 which is not possible (two angles cant be obtuse)
=> case a) rejected
=> all three factors positve
=> A,B,C <Ï€/2
=> acute
62asish is right..
sorry i goofed up in another post.. din see the solution completely but pinked it after a couple of steps :(
106k2 = -2k-1
=> (k+1)2 = 0
=> k = -1
so tanA + tanB + tanC = 1 = tanAtanBtanC
tanA +tanB+tanC ≥ 3 (iff tanA, tanB, tanC >0 )
but this is impossible
so one and only one of tanA, tanB and tanC is negative
so the triangle is obtuse
edit: courtesy soumya:
tanA+tan+BtanC = 1. This implies that atleast one in tanA, tanB, tanC is negative but since tanAtanBtanC > 0 so two out of tanA, tanB, tanC is negative which is not possible for any triangle..
so no triangle possible
341Shouldnt tan A + tan B + tan C = tan A tan B tan C for a triangle
But that would mean k2+1 = 0 here!
edit: thought i was goofing there
24hehe,,,ya,,,corrected now..
this is what happens when u reply after a looong time[3]
106tan takes negative value when angle >pi
tan is negative when angle is in 2nd quad.
