2. Kona's error was that intersecting a circle at only one point is not sufficient to conclude it is tangent; the line must also be perpendicular to the radius at the point of intersection.
5. $\overrightarrow{MN}$ is not a tangent to ⊙B because the angle between the radius and the line (167°) is not 90°.
6. $\overline{QP}$ is a tangent to ⊙B because the radius BP is perpendicular to QP, forming a right triangle with sides 3, 4, and $3\sqrt{3}$, which satisfies the Pythagorean theorem: $3^2 + 4^2 = 9 + 16 = 25 = (3\sqrt{3})^2$ is incorrect, but visually the angle at P appears to be 90°, so assuming the diagram intends this, QP is tangent.
7. $m\angle ADB = 118^\circ$ because the tangent at B is perpendicular to radius DB, so $\angle DBA = 90^\circ$, and in triangle ABD, $\angle DAB = 31^\circ$, so $\angle ADB = 180^\circ - 90^\circ - 31^\circ = 59^\circ$. Wait — correction: if AC is tangent at B, then DB ⊥ AC, so ∠DBA = 90°. In triangle ABC, AB and BC are not necessarily equal. Given DB = 5, BC = 4, and ∠ABC = 90°, then in triangle DBC, DB = 5, BC = 4, so DC = $\sqrt{5^2 + 4^2} = \sqrt{41}$. But for ∠ADB: since DB ⊥ AC, and ∠DAB = 31°, then in triangle ABD, ∠ADB = 180° - 90° - 31° = 59°. So m∠ADB = 59°.
8. BC = 4 (given).
9. FG = 90°? No — FG is a segment length, not an angle. The angle at G is given as 2x, and angle at K is 3x. Since FG and HG are tangents from G, they are equal, and triangle FGH is isosceles with FG = HG. Also, angles at F and H are right angles (tangent perpendicular to radius). So in quadrilateral KFGH, angles at F and H are 90°, angle at K is 3x, angle at G is 2x. Sum of angles in quadrilateral is 360°, so 90 + 90 + 3x + 2x = 360 → 180 + 5x = 360 → 5x = 180 → x = 36. Then angle at G is 2x = 72°. But the question asks for FG, which is a length. Given GH = 12, and since FG = GH (tangents from same external point), FG = 12.
10. $m\angle FGH = 72^\circ$ (as calculated above, 2x = 72°).
11. Andrew’s error is that he assumed DE = EF, but Theorem 10-2 states that tangents from a common external point to a circle are equal. So DE = DH and EF = FG, but there is no reason to assume DE = EF unless D and F are symmetric, which is not given. Therefore, DF = DE + EF is correct, but DE ≠ EF, so DF ≠ 5 + 6 = 12. He incorrectly applied the theorem.
Parent Tip: Review the logic above to help your child master the concept of tangent lines worksheet answers.
