Question Number 81983 by oyemi kemewari last updated on 17/Feb/20
Commented by jagoll last updated on 17/Feb/20
$${L}−{D}\:{or}\:{L}.{D}? \\ $$
Commented by oyemi kemewari last updated on 17/Feb/20
l-d
Commented by john santu last updated on 17/Feb/20
$$\mathrm{sin}\:\theta\:=\:\frac{{h}}{{L}}\:,\:\mathrm{cos}\:\theta\:=\:\frac{{D}}{{L}} \\ $$$$\Rightarrow{D}\:=\:{L}\:\mathrm{cos}\:\theta\: \\ $$$${L}−{D}\:=\:{L}−{L}\mathrm{cos}\:\theta\:=\:{L}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$=\:\frac{{L}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)}{\mathrm{1}+\mathrm{cos}\:\theta}\:=\:\frac{{L}\:\frac{{h}^{\mathrm{2}} }{{L}^{\mathrm{2}} }}{\mathrm{1}+\frac{{D}}{{L}}}\:=\:\frac{{h}^{\mathrm{2}} }{{L}}×\frac{{L}}{{L}+{D}} \\ $$$$=\:\frac{{h}^{\mathrm{2}} }{{L}+{D}}\: \\ $$$${tobe}\:{continue} \\ $$$$ \\ $$
Commented by mr W last updated on 17/Feb/20
$${question}\:{is}\:{simply}\:{wrong}! \\ $$$${L}−{D}=\frac{{h}^{\mathrm{2}} }{\mathrm{2}{L}}\:{can}\:{never}\:{be}\:{true}! \\ $$