[tex] \large \bold{SOLUTION:} [/tex]
We know that the measure of an external angle or outside angle of the circle, formed by two tangent segments, is half the difference of the measures of its intercepted arcs. Let [tex]C[/tex] be a point on the larger arc intercepted by an external angle 40°.
Given that [tex]\textsf{m}\overset{\Large \frown}{AB} = x\degree [/tex],
[tex] \begin{array}{l} \begin{aligned} \qquad 40\degree &= \frac{1}{2}(\textsf{m}\overset{\huge \frown}{ACB} - \textsf{m}\overset{\Large \frown}{AB}) \\ 40\degree &= \frac{1}{2}\big[(360 - x)\degree - x\degree\big] \\ 40\degree &= \frac{1}{2}(360 - 2x)\degree \\ 40 &= 180 - x \\ x &= 180 - 40 \\ x &= 140 \end{aligned} \\ \\ \implies \textsf{m}\overset{\Large \frown}{AB} = x\degree = \boxed{140\degree}\quad\textit{Answer} \end{array} [/tex]
[tex] \blue{\mathfrak{\#CarryOnLearning}} [/tex]
