Outer angles

The outer angle of the triangle ABC at the A vertex is 71°40 ' outer angle at the vertx B is 136°50'. What size has the inner triangle angle at the vertex C?

Result

C =  28.5 °

Solution:

$A = 180 - (71+40/60) = \dfrac{ 325 }{ 3 } \doteq 108.3333 \ ^\circ \ \\ B = 180 - (136+50/60) = \dfrac{ 259 }{ 6 } \doteq 43.1667 \ ^\circ \ \\ C = 180 - (A+B) = 180 - (108.3333+43.1667) = \dfrac{ 57 }{ 2 } = 28.5 = 28.5 ^\circ = 28^\circ 30'$

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