Question Number 203778 by mr W last updated on 27/Jan/24
Answered by mr W last updated on 27/Jan/24
Commented by mr W last updated on 27/Jan/24
$${an}\:{other}\:{way} \\ $$$${AD}//{EB} \\ $$$$\frac{{AD}}{{EB}}=\frac{\mathrm{3}}{\mathrm{3}+\mathrm{5}} \\ $$$$\Rightarrow{AD}=\frac{\mathrm{3}}{\mathrm{8}}×\mathrm{5}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$
Answered by esmaeil last updated on 27/Jan/24
$$\frac{\mathrm{1}}{\mathrm{2}}{AC}×{ADsin}\mathrm{60}+\frac{\mathrm{1}}{\mathrm{2}}{AB}×{ADsin}\mathrm{60}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{AB}×{ACsin}\mathrm{120}\rightarrow \\ $$$$\mathrm{3}{AD}+\mathrm{5}{AD}=\mathrm{15}\rightarrow{AD}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$
Commented by mr W last updated on 27/Jan/24
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Commented by esmaeil last updated on 27/Jan/24
$${thanks}. \\ $$$${you}\:{are}\:{wonderful}\:{i}\:{learned}\:{a}\:{lot} \\ $$$${from}\:{you}. \\ $$
Answered by ajfour last updated on 28/Jan/24
Commented by ajfour last updated on 28/Jan/24
$${here}\:\:\frac{{h}}{{k}}=\frac{\mathrm{5}}{\mathrm{3}},\:\:{b}=\mathrm{5},\:{c}=\mathrm{3} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}=\mathrm{25}+\mathrm{9}+\mathrm{15} \\ $$$${hence} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} =\mathrm{34}−\frac{\mathrm{16}}{\mathrm{4}}−\frac{\mathrm{30}}{\mathrm{64}}\left(\mathrm{25}+\mathrm{9}+\mathrm{15}\right) \\ $$$$\:\:\:=\mathrm{30}−\frac{\mathrm{15}}{\mathrm{32}}×\mathrm{49}\:=\frac{\mathrm{15}}{\mathrm{32}}\left(\mathrm{64}−\mathrm{49}\right) \\ $$$$\:\:\:=\frac{\mathrm{15}^{\mathrm{2}} }{\mathrm{32}} \\ $$$$\Rightarrow\:\:\:\:\:{s}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$
Commented by ajfour last updated on 28/Jan/24
