Given:
[tex]m\angle A=(2x-15)^\circ[/tex]
[tex]m\angle B=(x-5)^\circ[/tex]
[tex]\text{Exterior}\ m\angle C=148^\circ[/tex]
To find:
The [tex]m\angle ABC[/tex].
Solution:
In triangle ABC,
[tex]\text{Exterior}\ m\angle C=m\angle A+m\angle B[/tex] [Exterior angle theorem]
[tex]148=(2x-15)+(x-5)[/tex]
[tex]148=3x-20[/tex]
[tex]148+20=3x[/tex]
[tex]168=3x[/tex]
Divide both sides by 3.
[tex]\dfrac{168}{3}=x[/tex]
[tex]56=x[/tex]
Now,
[tex]m\angle ABC=(x-5)^\circ[/tex]
[tex]m\angle ABC=(56-5)^\circ[/tex]
[tex]m\angle ABC=51^\circ[/tex]
Therefore, [tex]m\angle ABC=51^\circ[/tex].
