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Question Number 70219 by Mr. K last updated on 02/Oct/19

Commented by MJS last updated on 02/Oct/19

$$a=54°$$$$b=101°$$$$c=126°$$$$d=79°$$$$\mathrm{I}\mathrm{calculated}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{vertices},$$$$\mathrm{all}\mathrm{the}\mathrm{side}\mathrm{lenghts}\mathrm{and}\mathrm{then}\mathrm{the}\mathrm{angles}\mathrm{of}$$$$\mathrm{the}\mathrm{triangles}\mathrm{with}\cos \alpha =\frac{a^2+b^2-c^2}{2ab}$$

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