A bird is sitting on the top, $T$, of a vertical pole 20 m above the ground and the angle of elevation of the bird at $T$ from a point $O$ on the ground is $45^\circ$. The bird takes one second to fly horizontally at a constant speed from $T$ to a point $B$ away from $O$, maintaining a constant height of 20 m above the ground throughout its flight. The angle of elevation of the bird at $B$ from $O$ is reduced to $30^\circ$. Find the speed of the bird in m/s. Answer ... m/s

1. Find the horizontal distance from O to the pole (let's call the base of the pole P).

Let $P$ be the point on the ground directly below $T$.
In $\triangle OPT$, $\angle OPT = 90^\circ$, $TP = 20$ m, and $\angle TOP = 45^\circ$.
$\tan 45^\circ = \frac{TP}{OP}$
$OP = \frac{20}{\tan 45^\circ} = 20$ m

2. Find the horizontal distance from O to the point directly below B (let's call it Q).

Let $Q$ be the point on the ground directly below $B$.
In $\triangle OQB$, $\angle OQB = 90^\circ$, $BQ = 20$ m, and $\angle BOQ = 30^\circ$.
$\tan 30^\circ = \frac{BQ}{OQ}$
$OQ = \frac{20}{\tan 30^\circ} = 20\sqrt{3}$ m

3. Calculate the distance flown and the speed.

Distance flown $TB = PQ = OQ - OP = 20\sqrt{3} - 20$ m.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{20\sqrt{3} - 20}{1} = 20(\sqrt{3} - 1) \approx 14.641$ m/s