In the given figure, angle ABC is formed by a tangent and a secant.
The angle formed by tangent and secant is given by
[tex]m\angle ABC=\frac{1}{2}(m\bar{AD}-m\bar{AC})[/tex]
Where mAD and mAC are the intercepted arcs.
For the given case,
[tex]\begin{gathered} m\angle ABC=(4x+15)\degree \\ m\bar{AD}=(17x+2)\degree \\ m\bar{AC}=(7x-10)\degree \end{gathered}[/tex]
Let us substitute the given values into the above formula and solve for x
[tex]\begin{gathered} m\angle ABC=\frac{1}{2}(m\bar{AD}-m\bar{AC}) \\ (4x+15)\degree=\frac{1}{2}\lbrack(17x+2)\degree-(7x-10)\degree\rbrack \\ 2\cdot(4x+15)\degree=(17x+2)\degree-(7x-10)\degree \\ 8x+30=17x+2-7x+10 \\ 8x-17x+7x=2+10-30 \\ -2x=-18 \\ x=\frac{-18}{-2} \\ x=9 \end{gathered}[/tex]
The value of x is 9
So, the measure of angle ABC is
[tex]\begin{gathered} m\angle ABC=4x+15 \\ m\angle ABC=4(9)+15 \\ m\angle ABC=36+15 \\ m\angle ABC=51\degree \end{gathered}[/tex]
Therefore, the measure of angle ABC is 51°
