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Question Number 47449 by ajfour last updated on 10/Nov/18

Commented by ajfour last updated on 10/Nov/18

$F i n d t h e a r e a o f t h e s h a d o w o f$ $t h e r i g h t c i r c u l a r c o n e (n o t$ $i n c l u d i n g i t s b a s e a r e a) .$ $S i s a p o i n t s o u r c e o f l i g h t, a t$ $a d i s t a n c e d f r o m c e n t e r o f$ $b a s e o f c o n e, a t a h e i g h t H .$
	Answered by mr W last updated on 10/Nov/18

	Commented by mr W last updated on 10/Nov/18

	$\frac{e + d}{e} = \frac{H}{h}$ $\Rightarrow e = \frac{h d}{H - h}$ $\cos θ = \frac{r}{e} = \frac{r (H - h)}{h d}$ $\Rightarrow θ = \cos^{- 1} \frac{r (H - h)}{h d}$ $\sin θ = \frac{\sqrt{h^{2} d^{2} - r^{2} {(H - h)}^{2}}}{h d}$ $A_{s h a d o w} = r e \sin θ - θ π = \frac{r h d}{H - h} \times \frac{\sqrt{h^{2} d^{2} - r^{2} {(H - h)}^{2}}}{h d} - θ r^{2}$ $\Rightarrow A_{s h a d o w} = \frac{r \sqrt{h^{2} d^{2} - r^{2} {(H - h)}^{2}}}{H - d} - r^{2} \cos^{- 1} \frac{r (H - h)}{h d}$
	Commented by ajfour last updated on 10/Nov/18

	$T h a n k y o u S i r .$
	Commented by mr W last updated on 10/Nov/18

	$t h a n k y o u f o r c h e c k i n g s i r!$ $c a n y o u f i g u r e i t o u t i f w e h a v e a$ $c y l i n d e r i n s t e a d o f a c o n e ?$
	Commented by ajfour last updated on 10/Nov/18

	Commented by ajfour last updated on 10/Nov/18

	$c a n t f i g u r e p r o p e r l y, n o t s u r e,$ $S i r .$
	Commented by MrW3 last updated on 11/Nov/18

	$i m a g e i s n i c e a n d c o r r e c t .$ $p l e a s e s o l v e i t s i r .$
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