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)=3253≐108.3333 ∘ B=180−(136+50/60)=2596≐43.1667 ∘ C=180−(A+B)=180−(108.3333+43.1667)=572=28.5=28.5∘=28∘30′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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