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Question Number 175476 by ajfour last updated on 31/Aug/22

Commented by ajfour last updated on 31/Aug/22

$A c o n e h a s a n i n s c r i b e d c u b e r e s t i n g$ $a t i t s b a s e . A t o p t h e c u b e i n s c r i b e d$ $i n t h e c o n e i s a s p h e r e o f r a d i u s r .$ $F i n d r i n t e r m s o f R, H ({c o n e}^{'} s$ $r a d i u s a n d h e i g h t) .$
	Answered by mr W last updated on 01/Sep/22

	Commented by mr W last updated on 01/Sep/22

	$a = e d g e l e n g t h o f c u b e$ $\tan θ = \frac{H}{R} = \frac{a}{R - \frac{a}{\sqrt{2}}}$ $\frac{H}{R} = \frac{a}{R - \frac{a}{\sqrt{2}}}$ $\Rightarrow a = \frac{R}{\frac{R}{H} + \frac{1}{\sqrt{2}}}$ $r = \frac{a}{\sqrt{2}} \tan \frac{θ}{2} = \frac{a}{\sqrt{2}} \times \frac{\frac{H}{\sqrt{H^{2} + R^{2}}}}{1 + \frac{R}{\sqrt{H^{2} + R^{2}}}}$ $\Rightarrow r = \frac{R H}{(\frac{\sqrt{2} R}{H} + 1) (\sqrt{H^{2} + R^{2}} + R)}$
	Commented by ajfour last updated on 01/Sep/22

	$w o w, f a n t a s t i c, t h a n k s S i r .$
	Commented by Tawa11 last updated on 02/Sep/22

	$Great sir$
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