The image shows a circle with center O and points N and K on the circumference. Lines are drawn from point M tangent to the circle at N and K. Given OM = 18 and ON = 9, find the angle NMK.

Solution:

• The triangle OMN is a right-angled triangle because the radius ON is perpendicular to the tangent MN at the point of tangency N.
• In △OMN, we have ON = 9 (radius) and OM = 18 (given).
• We can find ∠NMO using trigonometry. Since ∠OMN is opposite to the side ON, we use the sine function: sin(∠NMO) = ON / OM = 9 / 18 = 1/2.
• Therefore, ∠NMO = Arcsin(1/2) = 30°.
• Similarly, △OMK is a right-angled triangle, and ∠OMK = ∠NMO = 30°.
• The angle NMK is the sum of angles ∠NMO and ∠KMO. However, the diagram suggests that M is a single point and MN and MK are tangents from M to the circle. Therefore, ∠NMK is the angle formed by these two tangents.
• The triangle OMN is congruent to triangle OMK (RHS congruence: Right angle, Hypotenuse OM is common, ON = OK = radius).
• Therefore, ∠NMO = ∠KMO = 30°.
• The angle ∠NMK = ∠NMO + ∠KMO = 30° + 30° = 60°.

Answer: ∠NMK = 60°