Interior angles

In a quadrilateral ABCD, whose vertices lie on some circle, the angle at vertex A is 58 degrees and the angle at vertex B is 134 degrees. Calculate the sizes of the remaining interior angles.

Correct result:

C =  122 °
D =  46 °

Solution:

$$\begin{gathered} A=58^{\circ } \\ B=134^{\circ } \\ \mathrm{\angle }2DCB=DS{B}_1 \\ \mathrm{\angle }2DAB=DS{B}_2 \\ DS{B}_1+DS{B}_2=360^{\circ } \\ 2DCB+2DAB=360^{\circ } \\ DCB+DAB=180^{\circ } \\ C+A=180^{\circ } \\ A+C=180^{\circ } \\ C=180-A=180-58=122^{\circ } \end{gathered}$$

$$\begin{gathered} B+D=180^{\circ } \\ D=180-B=180-134=46^{\circ } \end{gathered}$$

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