Question Number 179515 by cortano1 last updated on 30/Oct/22
Answered by Acem last updated on 30/Oct/22
Commented by Acem last updated on 30/Oct/22
$$ \\ $$$$\:{D}=\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\Upsilon}\:=\:\sqrt{\mathrm{133}}\:\:\:;\:\Upsilon=\:\mathrm{120} \\ $$$$\:\alpha=\:\mathrm{arccos}\:\frac{{a}^{\mathrm{2}} +{D}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{aD}}\:=\:\mathrm{36}.\mathrm{58}° \\ $$$$\:\beta=\:\mathrm{arccos}\:\frac{{c}^{\mathrm{2}} +{D}^{\mathrm{2}} −{e}^{\mathrm{2}} }{\mathrm{2}{cD}}\:=\:\mathrm{42}.\mathrm{52}° \\ $$$$\:\gamma\:=\:\alpha+\beta=\:\mathrm{79}.\mathrm{1}° \\ $$$$\:\boldsymbol{{d}}=\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{ac}}\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\gamma}}\:=\:\mathrm{6} \\ $$$$\: \\ $$
Commented by Tawa11 last updated on 30/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cortano1 last updated on 30/Oct/22
$$\mathrm{nice} \\ $$