Solve Exercise 49 again using geometry rather than calculus. There is a unique circle passing through points $B$ and $C$ which is tangent to the street. Let $R$ be the point of tangency. Note that the two angles labeled $\psi$ in Figure 27 are equal because they subtend equal arcs on the circle. (a) Show that the maximum value of $\theta$ is $\theta=\psi$. Hint: Show that $\psi=\theta+\angle P B A$ where $A$ is the intersection of the circle with $P C$. (b) Prove that this agrees with the answer to Exercise 49 . (c) Show that $\angle Q R B=\angle R C Q$ for the maximal angle $\psi$.
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Step 1: Show that $\psi=\theta+\angle PBA$ where $A$ is the intersection of the circle with $PC$. Show more…
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