The angles of a triangle are ${(x-35)}^o,{(x-20)}^o$ and ${(x+40)}^o$. Find the three angles.

Given : Angles of $△$ $=(x-35)°,(x-20)°\text{and}(x+40)°$

$\because$ sum of angle of $△=180°$

$\therefore x-35°+x-20°+x+40°=180°$

$\therefore 3x-15°=180°$

$x=\frac{195°}{3}=65°$

$\therefore$ angles are $65°-35°=30°65°-20°=45°65°+40°=105°$