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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= (√(a^2 +b^2 −2ab cos Υ)) = (√(133)) ; Υ= 120 α= arccos ((a^2 +D^2 −b^2 )/(2aD)) = 36.58° β= arccos ((c^2 +D^2 −e^2 )/(2cD)) = 42.52° γ = α+β= 79.1° d= (√(a^2 +c^2 −2ac cos 𝛄)) = 6

$$ \\ $$$$\:{D}=\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\Upsilon}\:=\:\sqrt{\mathrm{133}}\:\:\:;\:\Upsilon=\:\mathrm{120} \\ $$$$\:\alpha=\:\mathrm{arccos}\:\frac{{a}^{\mathrm{2}} +{D}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{aD}}\:=\:\mathrm{36}.\mathrm{58}° \\ $$$$\:\beta=\:\mathrm{arccos}\:\frac{{c}^{\mathrm{2}} +{D}^{\mathrm{2}} −{e}^{\mathrm{2}} }{\mathrm{2}{cD}}\:=\:\mathrm{42}.\mathrm{52}° \\ $$$$\:\gamma\:=\:\alpha+\beta=\:\mathrm{79}.\mathrm{1}° \\ $$$$\:\boldsymbol{{d}}=\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{ac}}\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\gamma}}\:=\:\mathrm{6} \\ $$$$\: \\ $$

Commented by Tawa11 last updated on 30/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by cortano1 last updated on 30/Oct/22

nice

$$\mathrm{nice} \\ $$

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