The angles of a triangle sum up to [tex] 180^\circ [/tex]. So, if you know two of them (let's say [tex] \alpha [/tex] and [tex] \beta [/tex]), you can find the third angle [tex] \gamma [/tex] by subtraction:
[tex] \alpha +\beta +\gamma = 180^\circ \iff \gamma = 180^\circ-\alpha -\beta = 180^\circ-(\alpha +\beta) [/tex]
So, in the first case, the given angles sum to [tex] 120^\circ [/tex], so the remaining one must be [tex] 60^\circ [/tex]
In the second case, the given angles sum to [tex] 19^\circ [/tex], so the remaining one must be [tex] 161^\circ [/tex]
In the third case, the given angles sum to [tex] 108^\circ [/tex], so the remaining one must be [tex] 72^\circ [/tex]
In the fourth case, the given angles sum to [tex] 159^\circ [/tex], so the remaining one must be [tex] 21^\circ [/tex]
Finally, the triangle with two sides of the same length is the one with two angles of the same measure, so the third one, which is isosceles.
