Question Number 47449 by ajfour last updated on 10/Nov/18 | ||
Commented by ajfour last updated on 10/Nov/18 | ||
$${Find}\:{the}\:{area}\:{of}\:{the}\:{shadow}\:{of} \\ $$$${the}\:{right}\:{circular}\:{cone}\:\left({not}\right. \\ $$$$\left.{including}\:{its}\:{base}\:{area}\right). \\ $$$${S}\:{is}\:{a}\:{point}\:{source}\:{of}\:{light},\:{at} \\ $$$${a}\:{distance}\:{d}\:{from}\:{center}\:{of} \\ $$$${base}\:{of}\:{cone},\:{at}\:{a}\:{height}\:{H}. \\ $$ | ||
Answered by mr W last updated on 10/Nov/18 | ||
Commented by mr W last updated on 10/Nov/18 | ||
$$\frac{{e}+{d}}{{e}}=\frac{{H}}{{h}} \\ $$$$\Rightarrow{e}=\frac{{hd}}{{H}−{h}} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}}{{e}}=\frac{{r}\left({H}−{h}\right)}{{hd}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{{r}\left({H}−{h}\right)}{{hd}} \\ $$$$\mathrm{sin}\:\theta=\frac{\sqrt{{h}^{\mathrm{2}} {d}^{\mathrm{2}} −{r}^{\mathrm{2}} \left({H}−{h}\right)^{\mathrm{2}} }}{{hd}} \\ $$$${A}_{{shadow}} ={re}\:\mathrm{sin}\:\theta−\theta\pi=\frac{{rhd}}{{H}−{h}}×\frac{\sqrt{{h}^{\mathrm{2}} {d}^{\mathrm{2}} −{r}^{\mathrm{2}} \left({H}−{h}\right)^{\mathrm{2}} }}{{hd}}−\theta{r}^{\mathrm{2}} \\ $$$$\Rightarrow{A}_{{shadow}} =\frac{{r}\sqrt{{h}^{\mathrm{2}} {d}^{\mathrm{2}} −{r}^{\mathrm{2}} \left({H}−{h}\right)^{\mathrm{2}} }}{{H}−{d}}−{r}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \frac{{r}\left({H}−{h}\right)}{{hd}} \\ $$ | ||
Commented by ajfour last updated on 10/Nov/18 | ||
$${Thank}\:{you}\:{Sir}.\: \\ $$ | ||
Commented by mr W last updated on 10/Nov/18 | ||
$${thank}\:{you}\:{for}\:{checking}\:{sir}! \\ $$$${can}\:{you}\:{figure}\:{it}\:{out}\:{if}\:{we}\:{have}\:{a} \\ $$$${cylinder}\:{instead}\:{of}\:{a}\:{cone}? \\ $$ | ||
Commented by ajfour last updated on 10/Nov/18 | ||
Commented by ajfour last updated on 10/Nov/18 | ||
$${cant}\:{figure}\:{properly},\:{not}\:{sure}, \\ $$$${Sir}. \\ $$ | ||
Commented by MrW3 last updated on 11/Nov/18 | ||
$${image}\:{is}\:{nice}\:{and}\:{correct}. \\ $$$${please}\:{solve}\:{it}\:{sir}. \\ $$ | ||