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A cube of side 8 cm stands on a horizontal table. A hollow cone of height 20 cm is placed over the cube so that it rests on the table and touches the top four corners of the cube. Find the vertical angle of the cone.

I can't seem to figure out how to work out the length of the base, which seems to be necessary to work out the angle.

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  • $\begingroup$ The triangle above the square is similar to the big triangle. So, if you can find the angles of that triangle they will be the same as for the big triangle. $\endgroup$
    – Doug M
    May 5, 2017 at 15:10

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The picture you have presented is not accurate. enter image description here

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  • $\begingroup$ It is accurate. It's just that a line has been omitted. $\endgroup$ May 5, 2017 at 15:53
  • $\begingroup$ @ParclyTaxel, Sorry, I mean not in scale. $\endgroup$
    – Seyed
    May 5, 2017 at 16:07
  • $\begingroup$ I don't need accurate diagrams to solve this... $\endgroup$ May 5, 2017 at 16:08
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From the view in the question, the cube's width is $8\sqrt2$ cm, half of which (distance from face centre to corner on same) is $4\sqrt2$ cm. It is 12 cm from the top face to the apex, so the tangent of half the vertical angle is $\tan\theta=\frac{4\sqrt2}{12}=\frac{\sqrt2}3$. The full angle is then $$\tan^{-1}\frac{2\tan\theta}{1-\tan^2\theta}=\tan^{-1}\frac{2\sqrt2/3}{1-2/9}=\tan^{-1}\frac{6\sqrt2}7$$

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