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Question-14724




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=13,BC=16,AC=15  AD=DC,DE⊥AC    ..................................         DE=?
$${AB}=\mathrm{13},{BC}=\mathrm{16},{AC}=\mathrm{15} \\ $$$${AD}={DC},{DE}\bot{AC}\:\: \\ $$$$……………………………. \\ $$$$\:\:\:\:\:\:\:{DE}=? \\ $$
Answered by mrW1 last updated on 04/Jun/17
cos ∠C=((15^2 +16^2 −13^2 )/(2×15×16))=0.65  sin ∠C=(√(1−0.65^2 ))=0.7599    DE=CD×tan ∠C  =((15)/2)×((0.7599)/(0.65)) =8.768
$$\mathrm{cos}\:\angle{C}=\frac{\mathrm{15}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} −\mathrm{13}^{\mathrm{2}} }{\mathrm{2}×\mathrm{15}×\mathrm{16}}=\mathrm{0}.\mathrm{65} \\ $$$$\mathrm{sin}\:\angle{C}=\sqrt{\mathrm{1}−\mathrm{0}.\mathrm{65}^{\mathrm{2}} }=\mathrm{0}.\mathrm{7599} \\ $$$$ \\ $$$${DE}={CD}×\mathrm{tan}\:\angle{C} \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{0}.\mathrm{7599}}{\mathrm{0}.\mathrm{65}}\:=\mathrm{8}.\mathrm{768} \\ $$
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.
$${nice}\:{and}\:{smart}!\:{thank}\:{a}\:{lot}\:{my}\:{master}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17
cosC=((b^2 +a^2 −c^2 )/(2ab))  sinC=((2S)/(ab))  tgC=(((2S)/(ab))/((a^2 +b^2 −c^2 )/(2ab)))=((4S)/(a^2 +b^2 −c^2 ))  DE=CD.tgC=(b/2).((4S)/(a^2 +b^2 −c^2 ))=((2b.S)/(a^2 +b^2 −c^2 ))  DE=((2b.S)/(2ab.cosC))=(S/(a.cosC)).
$${cosC}=\frac{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$${sinC}=\frac{\mathrm{2}{S}}{{ab}} \\ $$$${tgC}=\frac{\frac{\mathrm{2}{S}}{{ab}}}{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}}=\frac{\mathrm{4}{S}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${DE}={CD}.{tgC}=\frac{{b}}{\mathrm{2}}.\frac{\mathrm{4}{S}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }=\frac{\mathrm{2}{b}.{S}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$${DE}=\frac{\mathrm{2}{b}.{S}}{\mathrm{2}{ab}.{cosC}}=\frac{{S}}{{a}.{cosC}}. \\ $$

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