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?

Correct result:

C =  28.5 °

Solution:

$$\begin{gathered} A=180-(71+40\mathrm{/}60)={\displaystyle \frac{325}{3}}\doteq 108.3333^{\circ } \\ B=180-(136+50\mathrm{/}60)={\displaystyle \frac{259}{6}}\doteq 43.1667^{\circ } \\ C=180-(A+B)=180-(108.3333+43.1667)={\displaystyle \frac{57}{2}}={{\displaystyle \frac{57}{2}}}^{\circ }=28.5^{\circ }=28^{\circ }30^{\mathrm{\prime }} \end{gathered}$$

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