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Math Help - HELPP !! 3D tRIG PROBLEM !

  1. #1
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    HELPP !! 3D tRIG PROBLEM !

    two straight roads PC, PD at right angles to one another, running up a hill ABCD which is inclined at an angle theta to the horizontal. The roads PC, PD are inclined at angles a,b to the horizontal. BC, AD are lines of greatest slope and AB is the horizontal; CE, DF are verticals from C,D to the horizontal plane.

    Show that sinsquaredtheta = sinsquareda + sinsquaredb

    AHH pLEASE HELP, im SO LOST ! T.T
    any help is GREATLLYYY APPRECIATED !!
    THANKYOU in advancee
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  2. #2
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    3D Trigonometry

    Hello iiharthero
    Quote Originally Posted by iiharthero View Post
    two straight roads PC, PD at right angles to one another, running up a hill ABCD which is inclined at an angle theta to the horizontal. The roads PC, PD are inclined at angles a,b to the horizontal. BC, AD are lines of greatest slope and AB is the horizontal; CE, DF are verticals from C,D to the horizontal plane.

    Show that sinsquaredtheta = sinsquareda + sinsquaredb

    AHH pLEASE HELP, im SO LOST ! T.T
    any help is GREATLLYYY APPRECIATED !!
    THANKYOU in advancee
    Suppose BC = AD = d. I assume that you've drawn a diagram. If so, you'll know that:

    CE = d \sin \theta, BE = d\cos\theta, PE = CE\cot a = d\sin\theta\cot a, PC = \frac{CE}{\sin a}=\frac{d\sin\theta}{\sin a}

    So PB^2 = PE^2-BE^2 = d^2\sin^2\theta \cot^2a -d^2\cos^2\theta

    which can be re-written in terms of \sin\theta and \sin a (which is what we want!) as \frac{d^2(\sin^2\theta-\sin^2a)}{\sin^2a}, using \cos^2\theta = 1 - \sin^2\theta and 1 + \cot^2a = \csc^2 a=\frac{1}{\sin^2a} (Can you do this bit?)

    Similarly, using the other side of the diagram:

     DP = \frac{d\sin\theta}{\sin b} and

    PA^2 = \frac{d^2(\sin^2\theta - \sin^2b)}{\sin^2b}

    Now for the bit where you need to keep a clear head!

    Using Pythagoras on \triangle DPC: DC^2 = AB^2= DP^2+PC^2

    \Rightarrow (PB+PA)^2 = PD^2 + PC^2

    \Rightarrow \frac{d^2(\sin^2\theta-\sin^2a)}{\sin^2a} + \frac{d^2(\sin^2\theta-\sin^2b)}{\sin^2b} + \frac{2d^2}{\sin a \sin b}\sqrt{(\sin^2\theta-\sin^2a)(\sin^2\theta - \sin^2b)} = \frac{d^2\sin^2\theta}{\sin^2b}+\frac{d^2\sin^2\th  eta}{\sin^2a}

    If you now multiply both sides by \sin^2a\sin^2b and divide by d^2 this will simplify to

    -2\sin a^2 \sin^2 b + 2\sin a\sin b\sqrt{(\sin^2\theta-\sin^2a)(\sin^2\theta - \sin^2b)}=0

    \Rightarrow \sin a \sin b = \sqrt{(\sin^2\theta-\sin^2a)(\sin^2\theta - \sin^2b)}, assuming \sin a\sin b \ne 0

    \Rightarrow \sin^2a \sin^2b = \sin^4\theta - \sin^2\theta(\sin^2 a + \sin^2 b) + \sin^2a\sin^2b

    \Rightarrow \sin^2\theta\Big(\sin^2\theta -(\sin^2a + \sin^2b)\Big) = 0

    \Rightarrow \sin^2\theta = \sin^2a+\sin^2b, again assuming \sin^2\theta \ne 0

    That wasn't so easy, was it?

    Grandad
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