AB is a diameter of a circle with centre O. A tangent is drawn at point A. C is a point on the circle such that BC produced meets the tangent at P . If ∠APC = 62° then find the measure of the minor arc AC.

The correct option is 28��.

Here is the step-by-step geometrical derivation to find the measure of the angle associated with the minor arc:

Step 1: Understand the properties of the tangent and diameter
We are given that $\mathrm{AB}$ is the diameter of a circle with centre $O$, and a tangent is drawn at point $A$ which meets the line $\mathrm{BC}$ produced at point $P$.
Since the diameter at the point of contact is perpendicular to the tangent line:

$\angle \mathrm{BAP}=90°$

Step 2: Calculate the angle in triangle ABP
In the right-angled triangle $\mathrm{ABP}$, the sum of the interior angles is $180°$:

$\angle \mathrm{ABP}+\angle \mathrm{BAP}+\angle \mathrm{APB}=180°$

Given that $\angle \mathrm{APC}=62°$, and since the points $B$, $C$, and $P$ lie on the same straight line, we have:

$\angle \mathrm{APB}=\angle \mathrm{APC}=62°$

Substituting these values into the angle sum equation:

$\angle \mathrm{ABP}+90°+62°=180°$

$\angle \mathrm{ABP}+152°=180°$

$\angle \mathrm{ABP}=180°-152°=28°$

Step 3: Relate the angle to the minor arc AC
Since $B$, $C$, and $P$ are collinear, the angle $\angle \mathrm{ABP}$ is equivalent to the angle $\angle \mathrm{ABC}$:

$\angle \mathrm{ABC}=28°$
The angle subtended by the minor arc $\mathrm{AC}$ at the circumference of the circle is $\angle \mathrm{ABC}$, which measures $28°$.