Quadrilateral in circle

A quadrilateral is inscribed in the circle. Its vertices divide the circle in a ratio of 1:2:3:4. Find the sizes of its interior angles.

Correct answer:

A =  90 °
B =  126 °
C =  90 °
D =  54 °

Step-by-step explanation:

$$\begin{gathered} {\Phi }_1=\sphericalangle ASB;{\Phi }_2=\sphericalangle BSC;{\Phi }_3=\sphericalangle CSD;{\Phi }_4=\sphericalangle DSA \\ {\Phi }_1:{\Phi }_2:{\Phi }_3:{\Phi }_4=o_1:o_2:o_3:o_4=1:2:3:4 \\ x=\frac{360}{1+2+3+4}=36{}^{\circ } \\ {\Phi }_1=1\cdot x=1\cdot 36=36{}^{\circ } \\ {\Phi }_2=2\cdot x=2\cdot 36=72{}^{\circ } \\ {\Phi }_3=3\cdot x=3\cdot 36=108{}^{\circ } \\ {\Phi }_4=4\cdot x=4\cdot 36=144{}^{\circ } \\ A=\frac{180-{\Phi }_1}{2}+\frac{180-{\Phi }_4}{2}=\frac{180-36}{2}+\frac{180-144}{2}=90^{\circ } \end{gathered}$$

$B=\frac{180-{\Phi }_1}{2}+\frac{180-{\Phi }_2}{2}=\frac{180-36}{2}+\frac{180-72}{2}=126^{\circ }$

$C=\frac{180-{\Phi }_2}{2}+\frac{180-{\Phi }_3}{2}=\frac{180-72}{2}+\frac{180-108}{2}=90^{\circ }$

$$\begin{gathered} D=\frac{180-{\Phi }_3}{2}+\frac{180-{\Phi }_4}{2}=\frac{180-108}{2}+\frac{180-144}{2}=54=54^{\circ } \\ \text{ Verifying Solution: } \\ s=A+B+C+D=90+126+90+54=360{}^{\circ } \end{gathered}$$

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