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=180A=18058=122=122A=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=180B=180134=46=46B+D=180 ^\circ \ \\ \ \\ D=180-B=180-134=46=46 ^\circ



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