A tangent and a secant are drawn to a circle from an external point P. The angle at P measures 38 degrees and the near intercepted arc measures 64 degrees. Find the far intercepted arc.

Step-by-step solution

1. 1

   Recall the external-angle theorem. For any two lines meeting outside a circle — two secants, a tangent and a secant, or two tangents — the angle formed is half the difference of the two intercepted arcs:

   $\angle P=\frac{1}{2}\left(\text{far arc}-\text{near arc}\right)$

   The contrast worth remembering: an angle with its vertex on the circle is half a single arc, and one inside the circle is half the sum of two arcs. Vertex position decides the rule.

2. 2

   Identify which arc is which. The near arc is the one between the two intersection points closer to $P$ (here $64^{\circ}$); the far arc is the one on the opposite side of the circle. The far arc is always the larger of the two, so the answer must exceed $64^{\circ}$ — a useful check before computing.

3. 3

   Substitute the given values.

   $38=\frac{1}{2}\left(\text{far}-64\right)$

4. 4

   Solve for the unknown arc. Multiply both sides by 2, then add 64:

   $76=\text{far}-64\ \Longrightarrow\ \text{far}=140^{\circ}$

5. 5

   Verify the configuration is consistent. The far arc $140^{\circ}$ is indeed larger than the near arc $64^{\circ}$ ✓, and together they use $204^{\circ}$ of the circle, leaving $156^{\circ}$ for the two remaining arcs cut off by the tangent point — a positive amount, so the picture is geometrically possible. Substituting back: $\tfrac12(140-64)=\tfrac12(76)=38^{\circ}$ ✓.