Question Number 47932 by ajfour last updated on 17/Nov/18
Commented by ajfour last updated on 17/Nov/18
$${Q}.\mathrm{47824} \\ $$
Answered by ajfour last updated on 17/Nov/18
![cos θ = (R/( (√(d^2 +(H−h−R)^2 )))) r = Rsin θ tan α = ((H−h−R)/d) ((BM)/r) = (((((H−h−R))/(sin α))−Rcos θ)/(H/(sin α))) ∠SAG = 90°+α−θ ⇒ AG = Htan (θ−α) AM = Hcot α−Htan (θ−α) CM = Htan (180°−θ−α)−Hcot α GM = Hcot α ....(i) AC = AM+CM =2a 2a = H[tan (180°−θ−α)−tan (θ−α)] B≡(BM,Hcot α−d) ON = Htan (θ−α)+a−d eq. of shadow ellipse (x^2 /b^2 )+(([y−(Htan (θ−α)+a−d)]^2 )/a^2 )=1 substituting B≡(BM,Hcot α−d) we find b . Area of shadow = πab .](https://www.tinkutara.com/question/Q47934.png)
$$\mathrm{cos}\:\theta\:=\:\frac{{R}}{\:\sqrt{{d}^{\mathrm{2}} +\left({H}−{h}−{R}\right)^{\mathrm{2}} }} \\ $$$${r}\:=\:{R}\mathrm{sin}\:\theta \\ $$$$\mathrm{tan}\:\alpha\:=\:\frac{{H}−{h}−{R}}{{d}} \\ $$$$\frac{{BM}}{{r}}\:=\:\frac{\frac{\left({H}−{h}−{R}\right)}{\mathrm{sin}\:\alpha}−{R}\mathrm{cos}\:\theta}{\frac{{H}}{\mathrm{sin}\:\alpha}} \\ $$$$\angle{SAG}\:=\:\mathrm{90}°+\alpha−\theta \\ $$$$\Rightarrow\:\:{AG}\:=\:{H}\mathrm{tan}\:\left(\theta−\alpha\right) \\ $$$${AM}\:=\:{H}\mathrm{cot}\:\alpha−{H}\mathrm{tan}\:\left(\theta−\alpha\right) \\ $$$${CM}\:=\:{H}\mathrm{tan}\:\left(\mathrm{180}°−\theta−\alpha\right)−{H}\mathrm{cot}\:\alpha \\ $$$${GM}\:=\:{H}\mathrm{cot}\:\alpha\:\:\:\:….\left({i}\right) \\ $$$${AC}\:=\:{AM}+{CM}\:=\mathrm{2}{a} \\ $$$$\:\mathrm{2}{a}\:=\:{H}\left[\mathrm{tan}\:\left(\mathrm{180}°−\theta−\alpha\right)−\mathrm{tan}\:\left(\theta−\alpha\right)\right] \\ $$$${B}\equiv\left({BM},{H}\mathrm{cot}\:\alpha−{d}\right) \\ $$$${ON}\:=\:{H}\mathrm{tan}\:\left(\theta−\alpha\right)+{a}−{d} \\ $$$${eq}.\:{of}\:{shadow}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{\left[{y}−\left({H}\mathrm{tan}\:\left(\theta−\alpha\right)+{a}−{d}\right)\right]^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${substituting}\:{B}\equiv\left({BM},{H}\mathrm{cot}\:\alpha−{d}\right) \\ $$$${we}\:{find}\:{b}\:. \\ $$$${Area}\:{of}\:{shadow}\:=\:\pi{ab}\:. \\ $$
Commented by ajfour last updated on 17/Nov/18
$${This}\:{isn}'{t}\:{less}\:{complicated}\:{than} \\ $$$${the}\:{vector}\:{way}..\:{dear}\:{MrW}\:{Sir}. \\ $$
Commented by mr W last updated on 17/Nov/18
$${if}\:{complicated}\:{or}\:{not},\:{this}\:{is}\:{relative}. \\ $$$${all}\:{roads}\:{lead}\:{to}\:{rome}. \\ $$$${thank}\:{you}\:{for}\:{this}\:{solution}\:{sir}!\: \\ $$$${you}\:{have}\:{made}\:{every}\:{thing}\:{very}\:{clear}. \\ $$