This solution details how to find the measure of angle OBC when tangents are drawn to a circle from an external point.
Consider the right-angled triangle OBA.
Use trigonometry to find $\angle OAB$. The side OB is opposite to $\angle OAB$.
$ \sin(\angle OAB) = \frac{Opposite}{Hypotenuse} = \frac{OB}{AO} = \frac{3 \text{ cm}}{6 \text{ cm}} = \frac{1}{2} $
From the sine value, we find:
$ \angle OAB = 30^\circ $
Since AO bisects $\angle BAC$, the angle $\angle BAC = 2 \times \angle OAB$.
$ \angle BAC = 2 \times 30^\circ = 60^\circ $
In $\triangle ABC$, AB = AC (tangents from A), so it's an isosceles triangle.
The base angles are equal: $\angle ABC = \angle ACB$.
The sum of angles in $\triangle ABC$ is $180^\circ$.
$ \angle BAC + \angle ABC + \angle ACB = 180^\circ $
$ 60^\circ + 2 \times \angle ABC = 180^\circ $
$ 2 \times \angle ABC = 180^\circ - 60^\circ = 120^\circ $
$ \angle ABC = \frac{120^\circ}{2} = 60^\circ $
We know that the radius OB is perpendicular to the tangent AB.
$ \angle ABO = 90^\circ $
Angle ABO can be split into two parts: $\angle OBC$ and $\angle ABC$.
$ \angle ABO = \angle OBC + \angle ABC $
Substitute the known values:
$ 90^\circ = \angle OBC + 60^\circ $
Solve for $\angle OBC$:
$ \angle OBC = 90^\circ - 60^\circ = 30^\circ $
The measure of angle OBC is $30^\circ$.
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