[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
In triangle ABC
[tex] \purple{\rm :\longmapsto\:\angle A + \angle B = 125 \degree \: - - - (1) }[/tex]
[tex] \purple{\rm :\longmapsto\:\angle B + \angle C = 150 \degree \: - - - (2)}[/tex]
We know,
Sum of all interior angles of a triangle is supplementary.
[tex] \purple{\rm :\longmapsto\:\angle A + \angle B + \angle C = 180\degree }[/tex]
On adding equation (1) and (2), we get
[tex] \purple{\rm :\longmapsto\:\angle A + \angle B + \angle B + \angle C = 125\degree + 150 \degree \:}[/tex]
[tex] \purple{\rm :\longmapsto\:\angle A + \angle B + \angle C + \angle B = 275\degree \:}[/tex]
[tex] \purple{\rm :\longmapsto\:180\degree + \angle B = 275\degree \:}[/tex]
[tex] \purple{\rm :\longmapsto\:\angle B = 275\degree - 180\degree \:}[/tex]
[tex] \purple{\rm :\longmapsto\:\angle B = 95\degree \:}[/tex]
On substituting the value in equation (1) and (2), we get
[tex] \purple{\rm :\longmapsto\:\angle A + 95\degree = 125\degree }[/tex]
[tex] \purple{\rm :\longmapsto\:\angle A = 125\degree - 95\degree }[/tex]
[tex] \purple{\rm :\longmapsto\:\angle A = 30\degree }[/tex]
Also, from equation (2), we get
[tex] \purple{\rm :\longmapsto\:95\degree + \angle C = 150\degree }[/tex]
[tex] \purple{\rm :\longmapsto\:\angle C = 150\degree - 95\degree }[/tex]
[tex] \purple{\rm :\longmapsto\:\angle C = 55\degree }[/tex]
Hence,
[tex]\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{\angle A = 30\degree } \\ \\ &\sf{\angle B = 95\degree } \\ \\ &\sf{\angle C = 55\degree } \end{cases}\end{gathered}\end{gathered}[/tex]
