Problems

A thin circular path with diameter \(AB\) is laid on horizontal ground. A vertical flagpole is erected with its base at a point \(D\) on the diameter \(AB\). The angles of elevation of the top of the flagpole from \(A\) and \(B\) are \(\alpha\) and \(\beta\) respectively (both are acute). The point \(C\) lies on the circular path with \(DC\) perpendicular to \(AB\) and the angle of elevation of the top of the flagpole from \(C\) is \(\phi\). Show that \(\cot\alpha\cot \beta = \cot^2\phi\). Show that, for any \(p\) and \(q\), \[ \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) = \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q) .\] Deduce that, if \(p\) and \(q\) are positive and \( p+q \le \tfrac12 \pi\), then \[ \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \, \] and hence show that \(\phi \le \tfrac12(\alpha+\beta)\) when \( \alpha +\beta \le \tfrac12 \pi\,\).

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Solution:

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\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

For this question, you may use the following approximations, valid if \(\theta \) is small: \ \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-\theta^2/2\,\). A satellite \(X\) is directly above the point \(Y\) on the Earth's surface and can just be seen (on the horizon) from another point \(Z\) on the Earth's surface. The radius of the Earth is \(R\) and the height of the satellite above the Earth is \(h\).

1. Find the distance \(d\) of \(Z\) from \(Y\) along the Earth's surface.
2. If the satellite is in low orbit (so that \(h\) is small compared with \(R\)), show that $$d \approx k(Rh)^{1/2},$$ where \(k\) is to be found.
3. If the satellite is very distant from the Earth (so that \(R\) is small compared with \(h\)), show that $$d\approx aR+b(R^2/h),$$ where \(a\) and \(b\) are to be found.