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Question Number 81983 by oyemi kemewari last updated on 17/Feb/20

Commented by jagoll last updated on 17/Feb/20

L−D or L.D?

$${L}−{D}\:{or}\:{L}.{D}? \\ $$

Commented by oyemi kemewari last updated on 17/Feb/20

l-d

Commented by john santu last updated on 17/Feb/20

sin θ = (h/L) , cos θ = (D/L) ⇒D = L cos θ L−D = L−Lcos θ = L(1−cos θ) = ((L(1−cos^2 θ))/(1+cos θ)) = ((L (h^2 /L^2 ))/(1+(D/L))) = (h^2 /L)×(L/(L+D)) = (h^2 /(L+D)) tobe continue

$$\mathrm{sin}\:\theta\:=\:\frac{{h}}{{L}}\:,\:\mathrm{cos}\:\theta\:=\:\frac{{D}}{{L}} \\ $$$$\Rightarrow{D}\:=\:{L}\:\mathrm{cos}\:\theta\: \\ $$$${L}−{D}\:=\:{L}−{L}\mathrm{cos}\:\theta\:=\:{L}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$=\:\frac{{L}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)}{\mathrm{1}+\mathrm{cos}\:\theta}\:=\:\frac{{L}\:\frac{{h}^{\mathrm{2}} }{{L}^{\mathrm{2}} }}{\mathrm{1}+\frac{{D}}{{L}}}\:=\:\frac{{h}^{\mathrm{2}} }{{L}}×\frac{{L}}{{L}+{D}} \\ $$$$=\:\frac{{h}^{\mathrm{2}} }{{L}+{D}}\: \\ $$$${tobe}\:{continue} \\ $$$$ \\ $$

Commented by mr W last updated on 17/Feb/20

question is simply wrong! L−D=(h^2 /(2L)) can never be true!

$${question}\:{is}\:{simply}\:{wrong}! \\ $$$${L}−{D}=\frac{{h}^{\mathrm{2}} }{\mathrm{2}{L}}\:{can}\:{never}\:{be}\:{true}! \\ $$

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