If a cone sits on this sphere, what 's the cone's volume?

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Since \(AB\) is tangent to the sphere, \(\angle BAO = 90^\circ\). We are also given that \(OA=6\), and that the vertical angle of the cone is \(84^\circ\), so \(\angle OBA = 42^\circ\), giving us

Now \[ \tan 42^\circ = \frac{\text{$OA$}}{\text{$AB$}} \implies \text{$AB$} = \frac{6}{\tan 42^\circ} = \quantity{6.66}{cm}. \]

The length we require is \(AC\). If we consider the triangle \(CBA\), we have that \[\begin{align*} \sin 42^\circ = \frac{\text{$AC$}}{\text{$AB$}} \implies \text{$AC$} &= \frac{6 \sin 42^\circ}{\tan 42^\circ}\\ &= 6 \cos 42^\circ\\ &= \quantity{4.46}{cm}. \end{align*}\]

The point \(D\) is the one on the sphere closest to B, so our goal is to find \(BD\). We have \(OD=6\), since it’s a radius of the sphere, and hence \[ \text{$BD$} = \text{$BO$} - 6. \] So we need to calculate the length of \(BO\). Considering \(\triangle AOB\), we have that \[ \sin 42^\circ = \frac{\text{$AO$}}{\text{$BO$}} \implies \text{$BO$} = \frac{6}{\sin 42^\circ} \] and therefore \[ \text{$BD$} = \frac{6}{\sin 42^\circ} - 6= \quantity{2.97}{cm}. \]

The volume of a cone is \[ \frac{1}{3} \times \text{area of the base} \times \text{height of the cone}. \] For us, the base is a disc with \(AC\) as a radius, and the height of the cone is \(BC\). Considering \(\triangle ACB\), we see that \[ \tan 42^\circ = \frac{\text{$AC$}}{\text{$BC$}} \implies \text{$BC$} = \frac{\text{$AC$}}{\tan 42^\circ}. \] Thus the volume of the cone is \[\begin{align*} \frac{1}{3} \times \pi \times \text{$AC$}^2 \times \text{$BC$} &= \frac{\pi (6 \cos 42^\circ)^3}{3 \tan 42^\circ} \\ &= \frac{72 \pi (\cos 42^\circ)^4}{\sin 42^\circ}\\ &= \quantity{103.1}{cm^3}. \end{align*}\]