27) From the diagram in the question we can say that
From the above diagram we can say that (6x + 13) and (7x - 2) are adjacent and suplementary ( i.e they add up to 180 degrees), thus,
[tex]\begin{gathered} (6x+13)^0+(7x-2)^0^{}=180^0 \\ 6x^0+7x^0+13^0-2^0=180^0 \\ 13x^0+11^0=180^0 \\ 13x^0=169^0 \\ x=\frac{169}{13}=13 \end{gathered}[/tex]
The value of the angles are
[tex]\begin{gathered} (6x+13)^0=(78+13)^0=91^0 \\ (5x+4)^0=(65+4)^0=69^0 \\ (7x-2)^0=(91-2)^0=89^0 \end{gathered}[/tex]
Hence, angle ABC is 890
