In the given figure, $PA$ is tangent to a circle with centre $O$. If $\angle APO = 30°$ and $OA = 2.5$ cm, then $OP$ is equal to

In the given figure, $PA$ is tangent to a circle with centre $O$. If $\angle APO = 30°$ and $OA = 2.5$ cm, then $OP$ is equal to 

Given:
- $ \displaystyle PA $ is a tangent to the circle at point $ \displaystyle A $.
- $ \displaystyle O $ is the center of the circle.
- $ \displaystyle \angle APO = 30^\circ $.
- $ \displaystyle OA = 2.5 $ cm.

Since the radius is perpendicular to the tangent at the point of contact, we have:

$ \displaystyle \angle OAP = 90^\circ $.

In right-angled triangle $ \displaystyle OAP $, using the cosine function:

$ \displaystyle \cos \angle APO = \frac{OA}{OP} $

Substituting the values:

$ \displaystyle \cos 30^\circ = \frac{2.5}{OP} $

Since $ \displaystyle \cos 30^\circ = \frac{\sqrt{3}}{2} $, we get:

$ \displaystyle \frac{\sqrt{3}}{2} = \frac{2.5}{OP} $

Cross multiply:

$ \displaystyle OP = \frac{2.5 \times 2}{\sqrt{3}} $

$ \displaystyle OP = \frac{5}{\sqrt{3}} $

Rationalizing the denominator:

$ \displaystyle OP = \frac{5\sqrt{3}}{3} $ cm.

Thus, the length of $ \displaystyle OP $ is **$ \displaystyle \frac{5}{\sqrt{3}} $ cm**.

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