A person walking on a straight road towards a tower observes that the angle of elevation of the top of a flag staff on the tower is - after going a distance a towards the tower he finds that the flag staff subtends the greatest angle - Prove that the length of the flag staff is 2a sin-sin-cos- - cos - - -

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Question

A person walking on a straight road towards a tower observes that the angle of elevation of the top of a flag staff on the tower is $α$ ; after going a distance a towards the tower he finds that the flag staff subtends the greatest angle $β$ . Prove that the length of the flag staff is $\frac{2 a sin α sin β}{cos α - cos (α - β)}$ .

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Solution

Let OP and PQ represent the tower and flag staff respectively and let the circle through P, Q touch the road AO at B where A is the initial position of the person where the elevation of Q is $α$ . Then B is the point at which the flag staff subtends the greatest angle $β$ , Now if $∠ P B O = θ$ , then by geometry, $∠ P Q B = θ$ also.
Hence from $Δ O B Q, β + θ + θ = 90^{o}$ ...(1)
so that $2 θ = 90^{o} - β$ . Also $∠ A Q B = β + θ - α$ .
As in part (a) we shall apply sine rule on $Δ B Q P$ and $Δ B Q A$ and eliminate the common side BQ to get PQ in terms of $α$ .
Note from $Δ A B Q$ ,
$\frac{B Q}{sin α} = \frac{1}{sin (β + θ - α)}$ ...(2)
and from $∠ B P Q$ ,
$\frac{B Q}{sin (π (β + θ))} = \frac{P Q}{sin β}$
Hence from (2) and (3), on equating the value of BQ, we get
$B Q = \frac{a sin α}{sin (β + θ - α)} = \frac{sin (β + θ)}{sin β} P Q$
or $P Q = \frac{a sin α sin β}{sin (β + θ - α) sin (β + θ)}$
$= \frac{2 a sin α sin β}{2 sin (β + θ - α) sin (β + θ)}$
$= \frac{2 a sin α sin β}{cos α - cos (2 β + 2 θ - α)}$
$= \frac{2 a sin α sin β}{cos α - cos (α - β)}$ by (1)

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