In the given figure, 𝐴𝑂𝐵 is a diameter of the circle with the centre 𝑂 and 𝐴𝐶 is a tangent to the circle at 𝐴. If ∠𝐵𝑂𝐶 = 130°, then find ∠𝐴𝐶𝑂.

In the given figure, ∠𝐵𝑂𝐶 is an external angle for ∠𝑂𝐴𝐶 and ∠𝐴𝐶𝑂. 

By the exterior angle property, we know that the exterior angle of a triangle is equal to the sum of its two corresponding interior opposite angles.

⇒ ∠𝐵𝑂𝐶 =∠𝑂𝐴𝐶 + ∠𝐴𝐶𝑂

Since radius of a circle is always proportional to the tangent at any point on it, ∠𝑂𝐴𝐶 = 90°,

⇒ 130°= 90°+ ∠𝐴𝐶𝑂

⇒ ∠𝐴𝐶𝑂 = 40°

Hence, the angle ∠𝐴𝐶𝑂 will be 40°.