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Thread: trigonometry (2)

  1. #1
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    trigonometry (2)

    Two lines , at right angle to one another , lie on a plane which is inclined at angle $\displaystyle \theta $ to the horizontal . If the 2 lines intersect at a point on the horizontal and their angles of inclination to the horizontal are $\displaystyle \alpha $ and $\displaystyle \beta $ respectively , prove that

    $\displaystyle \sin^2 \theta=\sin^2 \alpha + \sin^2 \beta$
    Last edited by mr fantastic; Oct 7th 2009 at 12:59 AM. Reason: Fixed latex
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Two lines , at right angle to one another , lie on a plane which is inclined at angle $\displaystyle \theta $ to the horizontal . If the 2 lines intersect at a point on the horizontal and their angles of inclination to the horizontal are $\displaystyle \alpha $ and $\displaystyle \beta $ respectively , prove that

    $\displaystyle \sin^2 \theta=\sin^2 \alpha + \sin^2 \beta$
    Drawing a good 3-D diagram is a great help here, so do the following:

    • Draw a wedge with a horizontal rectangular base PQRS, and a sloping rectangular face PABQ, so that the line AB is parallel to PS, with A vertically above Q and B vertically above R.

      Then the angle that the sloping face makes with the horizontal $\displaystyle = \angle APQ = \angle BSR = \theta$.


    • Now mark a point O on PS. Join O to A, B, Q, R. Suppose that OA and OB are the two perpendicular lines referred to in the question. Then the angles these lines make with the horizontal are:

      $\displaystyle \angle AOQ = \alpha$ and $\displaystyle \angle BOR = \beta$

    Then all the triangles are right-angled (except $\displaystyle \triangle$ ROQ). Suppose that $\displaystyle AQ = BR = h$. Then we have:

    $\displaystyle AP = \frac{h}{\sin\theta},\, OA = \frac{h}{\sin\alpha}$

    $\displaystyle \Rightarrow OP = \sqrt{\frac{h^2}{\sin^2\alpha}-\frac{h^2}{\sin^2\theta}}$

    Similarly $\displaystyle OS = \sqrt{\frac{h^2}{\sin^2\beta}-\frac{h^2}{\sin^2\theta}}$

    $\displaystyle \Rightarrow PS^2 = (PO+OS)^2$

    $\displaystyle = \Bigg(\sqrt{\frac{h^2}{\sin^2\alpha}-\frac{h^2}{\sin^2\theta}}+\sqrt{\frac{h^2}{\sin^2\ beta}-\frac{h^2}{\sin^2\theta}}\Bigg)^2$

    But $\displaystyle PS^2=AB^2=OA^2+OB^2$

    $\displaystyle \Rightarrow \Bigg(\sqrt{\frac{h^2}{\sin^2\alpha}-\frac{h^2}{\sin^2\theta}}+\sqrt{\frac{h^2}{\sin^2\ beta}-\frac{h^2}{\sin^2\theta}}\Bigg)^2=\frac{h^2}{\sin^ 2\alpha}+\frac{h^2}{\sin^2\beta}$

    Multiply through by $\displaystyle \sin^2\alpha\sin^2\beta\sin^2\theta$, and divide by $\displaystyle h^2$:

    $\displaystyle \Big(\sin\beta\sqrt{\sin^2\theta - \sin^2\alpha}+ \sin\alpha\sqrt{\sin^2\theta - \sin^2\beta}\Big)^2=\sin^2\theta(\sin^2\alpha + \sin^2\beta)$

    $\displaystyle \Rightarrow \sin^2\beta(\sin^2\theta - \sin^2\alpha) + 2\sin\alpha\sin\beta\sqrt{(\sin^2\theta - \sin^2\alpha)(\sin^2\theta - \sin^2\beta)}$ $\displaystyle +\sin^2\alpha(\sin^2\theta - \sin^2\beta)=$ $\displaystyle \sin^2\theta(\sin^2\alpha + \sin^2\beta)$

    $\displaystyle \Rightarrow 2\sin\alpha\sin\beta\sqrt{(\sin^2\theta - \sin^2\alpha)(\sin^2\theta - \sin^2\beta)} - 2\sin^2\alpha\sin^2\beta = 0$

    $\displaystyle \Rightarrow \sqrt{(\sin^2\theta - \sin^2\alpha)(\sin^2\theta - \sin^2\beta)} =\sin^2\alpha\sin^2\beta $, assuming $\displaystyle \sin\alpha\sin\beta \ne 0$

    $\displaystyle \Rightarrow \sin^4\theta-\sin^2\theta(\sin^2\alpha +\sin^2\beta)=0$

    $\displaystyle \Rightarrow \sin^2\theta = \sin^2\alpha+\sin^2\beta$, assuming $\displaystyle \sin\theta \ne 0$

    Phew!

    Grandad
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