Question Number 213119 by a.lgnaoui last updated on 30/Oct/24
$$\mathrm{determiner}\:\boldsymbol{\mathrm{a}}\:\mathrm{et}\:\boldsymbol{\mathrm{b}}\:\mathrm{par}\:\:\:;\:\:\mathrm{AB}\:\bot\mathrm{BC} \\ $$$$\begin{cases}{\mathrm{AM}\:=\mathrm{5}}\\{\mathrm{AC}\:\:=\mathrm{16}}\end{cases} \\ $$
Commented by a.lgnaoui last updated on 30/Oct/24
Answered by A5T last updated on 30/Oct/24
$${AN}×\mathrm{16}=\mathrm{5}×{AB}\Rightarrow\left(\mathrm{16}−{b}\right)\mathrm{16}=\mathrm{5}\left(\mathrm{5}+{a}\right) \\ $$$$\frac{{sin}\mathrm{30}}{{BC}}=\frac{\mathrm{1}}{\mathrm{16}}\Rightarrow{BC}=\mathrm{8}\Rightarrow\mathrm{5}+{a}=\sqrt{\mathrm{16}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{a}=\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{5}\Rightarrow{b}=\mathrm{16}−\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by a.lgnaoui last updated on 30/Oct/24
$$\mathrm{thanks}\:\mathrm{there}\:\mathrm{i}\:\mathrm{2}\:\mathrm{methodes} \\ $$$$\mathrm{for}\:\mathrm{calcul}\:\left(\mathrm{methode}\:\mathrm{du}\:\mathrm{cosinus}\:\mathrm{et}\:\mathrm{secantes}\right) \\ $$
Answered by MrGaster last updated on 30/Oct/24
$${AM}^{\mathrm{2}} +{M}\mathrm{C}^{\mathrm{2}} =\mathrm{AC}^{\mathrm{2}} \\ $$$$\mathrm{5}^{\mathrm{2}} +{M}\mathrm{C}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} \\ $$$${MC}^{\mathrm{2}} =\mathrm{256}−\mathrm{25} \\ $$$${M}\mathrm{C}^{\mathrm{2}} =\mathrm{231} \\ $$$${MC}=\sqrt{\mathrm{231}} \\ $$$${AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} ={AC}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{256} \\ $$$${a}={AM}=\mathrm{5} \\ $$$$\mathrm{5}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{256} \\ $$$${b}^{\mathrm{2}} =\mathrm{256}−\mathrm{25} \\ $$$${b}^{\mathrm{2}} =\mathrm{231} \\ $$$${b}=\sqrt{\mathrm{231}} \\ $$
Commented by A5T last updated on 30/Oct/24
$${AM}^{\mathrm{2}} +{MC}^{\mathrm{2}} \neq{AC}^{\mathrm{2}} \\ $$
Commented by MrGaster last updated on 30/Oct/24
Thanks for correcting me!