Question Number 70219 by Mr. K last updated on 02/Oct/19
Commented by MJS last updated on 02/Oct/19
$${a}=\mathrm{54}° \\ $$$${b}=\mathrm{101}° \\ $$$${c}=\mathrm{126}° \\ $$$${d}=\mathrm{79}° \\ $$$$\mathrm{I}\:\mathrm{calculated}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vertices}, \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lenghts}\:\mathrm{and}\:\mathrm{then}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{cos}\:\alpha\:=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$