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Math Help - Trigonometry Problem: BP1 = BP2

  1. #1
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    Trigonometry Problem: BP1 = BP2



    How do you resolve this one?
    What i got till now:

    BP1= 16+12/tanx
    BP2= 12+16tanx

    -----------------------

    BP1=BP2 = 16+12/tanx = 12+16tanx = (...) = 4tanx - 16tanx2 +12=0


    Is it correct?
    All you have to do is resolve the equation, find tanx, then sinx and cosx and subsitute them in C(x)?


    Thanks in advance!
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  2. #2
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    Re: Trigonometry Problem: BP1 = BP2

    I think that you're making this more complicated than it has to be. BP1=BP2. Can't you just use this to show that tanx=1?
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Trigonometry Problem: BP1 = BP2

    If \angle B = 90 ^{\circ} then \tan{x}=\frac{BP_2}{BP_1}=1 because it is given that BP_1=BP_2.

    Find \angle x.


    \tan{x}=1

    \Rightarrow \sin{x}=\cos{x}

    \Rightarrow x=45 ^{\circ}

    Find P_1 A
    \sin{x}=\frac{12}{P_1 A} ...................( \sin{x}=\frac{\text{Perpendicular}}{\text{ Hypotenuse }})

    \Rightarrow \sin{45^{\circ}}=\frac{12}{P_1 A}

    \Rightarrow P_1 A=12\sqrt{2}

    Find P_2 A
    \angle B+x+\angle P_2 =180 ^{\circ} .......................(Angle Sum Property)

    \Rightarrow \90^{\circ}+45^{\circ}+\angle P_2 =180 ^{\circ}

    \Rightarrow \angle P_2 =45^{\circ}

    \sin{\angle P_2 }=\frac{16}{P_2 A} ...................( \sin{x}=\frac{\text{Perpendicular}}{\text{ Hypotenuse }})

    \Rightarrow \sin{45^{\circ}}=\frac{16}{P_2 A}

    \Rightarrow P_2 A=16\sqrt{2}

    Find P_1 P_2
    P_1 P_2 = P_1 A + P_2 A =12\sqrt{2}+16\sqrt{2}=28\sqrt{2}

    Did you understand? In fact you could also solve this question without the use of trigonometry.
    Last edited by sbhatnagar; November 13th 2011 at 04:14 AM.
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