Question

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

• A. 2.5 cm
• B. 5 cm
• C. \(\frac{5}{\sqrt{3}}\) cm
• D. 2 cm

Answer 1

The Correct Option is B

Given:
- \(PA\) is tangent to the circle at point \(A\).
- \(OA\) is the radius of the circle.
- \(\angle APO = 30^\circ\), \(OA = 2.5\, \text{cm}\).

Step 1: Understand tangent-radius property
- Radius drawn to the tangent at point of contact is perpendicular to the tangent.
- So, \(\angle OAP = 90^\circ\).

Step 2: Consider right-angled triangle \(\triangle OAP\)
- Using \(\sin\) of angle \(APO\):
\[ \sin(\angle APO) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{OA}{OP} \] \[ \sin 30^\circ = \frac{2.5}{OP} \]

Step 3: Use value of \(\sin 30^\circ\)
\[ \sin 30^\circ = \frac{1}{2} \] Substitute:
\[ \frac{1}{2} = \frac{2.5}{OP} \]

Step 4: Solve for \(OP\)
Cross-multiply:
\[ OP = 2 \times 2.5 = 5 \, \text{cm} \]

Final Answer:
\[ \boxed{5\, \text{cm}} \]