Question Number 14724 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 03/Jun/17
$${AB}=\mathrm{13},{BC}=\mathrm{16},{AC}=\mathrm{15} \\ $$$${AD}={DC},{DE}\bot{AC}\:\: \\ $$$$……………………………. \\ $$$$\:\:\:\:\:\:\:{DE}=? \\ $$
Answered by mrW1 last updated on 04/Jun/17
$$\mathrm{cos}\:\angle{C}=\frac{\mathrm{15}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} −\mathrm{13}^{\mathrm{2}} }{\mathrm{2}×\mathrm{15}×\mathrm{16}}=\mathrm{0}.\mathrm{65} \\ $$$$\mathrm{sin}\:\angle{C}=\sqrt{\mathrm{1}−\mathrm{0}.\mathrm{65}^{\mathrm{2}} }=\mathrm{0}.\mathrm{7599} \\ $$$$ \\ $$$${DE}={CD}×\mathrm{tan}\:\angle{C} \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{0}.\mathrm{7599}}{\mathrm{0}.\mathrm{65}}\:=\mathrm{8}.\mathrm{768} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17
$${nice}\:{and}\:{smart}!\:{thank}\:{a}\:{lot}\:{my}\:{master}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17
$${cosC}=\frac{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$${sinC}=\frac{\mathrm{2}{S}}{{ab}} \\ $$$${tgC}=\frac{\frac{\mathrm{2}{S}}{{ab}}}{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}}=\frac{\mathrm{4}{S}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${DE}={CD}.{tgC}=\frac{{b}}{\mathrm{2}}.\frac{\mathrm{4}{S}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }=\frac{\mathrm{2}{b}.{S}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${DE}=\frac{\mathrm{2}{b}.{S}}{\mathrm{2}{ab}.{cosC}}=\frac{{S}}{{a}.{cosC}}. \\ $$