, 2012

Since $AB=AC$, then $\triangle ABC$ is isosceles and $\angle ABC = \angle ACB$. Note that $\angle BAC = \theta$.

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The angles in $\triangle ABC$ add to $180^\circ$, so $\angle ABC + \angle ACB + \angle BAC = 180^\circ$.

Thus, $2\angle ACB + \theta = 180^\circ$ or $\angle ABC = \angle ACB = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{1}{2}\theta$.

Now $\triangle BCD$ is isosceles as well with $BC=BD$ and so $\angle CDB = \angle DCB = 90^\circ - \frac{1}{2}\theta$.

Since the angles in $\triangle BCD$ add to $180^\circ$, then $$\angle CBD = 180^\circ - \angle DCB - \angle CDB = 180^\circ - (90^\circ - \tfrac{1}{2}\theta) - (90^\circ - \tfrac{1}{2}\theta) = \theta$$ Now $\angle EBD = \angle ABC - \angle DBC = (90^\circ - \frac{1}{2}\theta) - \theta = 90^\circ - \frac{3}{2}\theta$.

Since $BE = ED$, then $\angle EDB = \angle EBD = 90^\circ - \frac{3}{2}\theta$.

Therefore, $\angle BED = 180^\circ - \angle EBD - \angle EDB = 180^\circ - (90^\circ - \frac{3}{2}\theta) - (90^\circ - \frac{3}{2}\theta) = 3\theta$.
Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.

Since the radius of the ferris wheel is 9 m (half of the diameter of 18 m) and $B$ is 1 m above the ground, then $O$ is $9+1=10$ m above the ground.

Let $\angle TOP = \theta$.

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Since the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\angle TOQ = 2\theta$.

Draw a perpendicular from $P$ to $R$ on $TB$ and from $Q$ to $G$ on $TB$.

Since $P$ is 16 m above the ground and $O$ is 10 m above the ground, then $OR = 6$ m.

Since $OP$ is a radius of the circle, then $OP = 9$ m.

Looking at right-angled $\triangle ORP$, we see that $\cos \theta = \dfrac{OR}{OP} = \dfrac{6}{9}=\dfrac{2}{3}$.

Since $\cos \theta = \frac{2}{3} < \frac{1}{\sqrt{2}} = \cos (45^\circ)$, then $\theta > 45^\circ$.

This means that $2\theta > 90^\circ$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.

Since $\angle TOQ = 2\theta$, then $\angle QOG = 180^\circ - 2\theta$.

Kolapo’s height above the ground at $Q$ equals 1 m plus the length of $BG$.

Now $BG = OB - OG$. We know that $OB = 9$ m.

Also, considering right-angled $\triangle QOG$, we have $$OG = OQ \cos(\angle QOG) = 9\cos(180^\circ - 2\theta) = -9\cos(2\theta) = -9(2\cos^2\theta - 1)$$ Since $\cos \theta = \frac{2}{3}$, then $OG = -9(2(\tfrac{2}{3})^2 - 1) = -9(\tfrac{8}{9}-1) = 1$ m.

Therefore, $BG = 9 - 1 = 8$ m and so $Q$ is $1+8=9$ m above the ground.