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[Trignometry]Crack me If you can
  • sin 2a+sin2b +sin2c+4cos(a )cos (b)=4

    where a b c are the angles of triangle. But type of triangle can be.
  • 5 Comments sorted by
  • Do you mean "What type of triangle can it be " ???
  • i mean aditi, is it:

    Equilateral, right,actute, obtuse or an isosceles Trangle :-)

    Hope my question is clear now! :-)
  • sin2A+sin2B+sin2C+4cosAcosB=4



    sin2A+sin2B+sin2C+2cos(A+B)+2cos(A-B)=4



    2sin(A+B)cos(A-B)+2sinCcosC-2cosC+2cos(A-B)=4



    cos(A-B)[sinC+1]+cosC[sinC-1]=2



    if this is true then there are two cases:



    (1)either one of them is equal to 2 therefore,cos(A-B)[sinC+1]=2



    which means that sinC=1 therefore, the triangle is right-angled.



    (2)both of them equal to 1 but sinC is always less than or equal to one.if sinC=1 the value becomes "zero" or if sinC is less than one the value becomes negative.



    therefore only first case is valid.
  • @Saurabh ... Cool solution
    I could have never thought about it this way !!
    Do you think this kind of question can be asked in AIEEE ?

    Also , could you put up some more questions like these ...more analytical than numerical ??
  • hey great attempt saurabh!!

    answer is right triangle and isosceles

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