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.

Result

C =  122 °
D =  46 °

Solution:

A=58 ∘ B=134 ∘  ∠2DCB=DSB1 ∠2DAB=DSB2 DSB1+DSB2=360∘ 2DCB+2DAB=360∘ DCB+DAB=180∘ C+A=180∘  A+C=180∘  C=180−A=180−58=122=122∘A=58 \ ^\circ \ \\ B=134 \ ^\circ \ \\ \ \\ ∠ 2DCB=DSB_{ 1 } \ \\ ∠ 2DAB=DSB_{ 2 } \ \\ DSB_{ 1 } + DSB_{ 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=122 ^\circ
B+D=180∘  D=180−B=180−134=46=46∘B+D=180 ^\circ \ \\ \ \\ D=180-B=180-134=46=46 ^\circ

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