# trigonometry (2)

• Oct 7th 2009, 12:35 AM
thereddevils
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

$\displaystyle \sin^2 \theta=\sin^2 \alpha + \sin^2 \beta$
• Oct 7th 2009, 01:49 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
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!