Question

In the given figure, PA and PB are tangents to a circle centred at O. If \(\angle OAB = 15^\circ\), then \(\angle APB\) equals :

• A. 30^
• B. 15^
• C. 45^
• D. 10^

Hint:

A highly useful direct shortcut relation for this specific standard geometry configuration is:
\( \angle APB = 2 \cdot \angle OAB \).
Using this shortcut: \( \angle APB = 2 \times 15^\circ = 30^\circ \).
This direct formula can save a lot of time in competitive and board exams!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question is based on the properties of tangents drawn from an external point to a circle.
We are given that \(PA\) and \(PB\) are tangents, \(O\) is the center of the circle, and \(\angle OAB = 15^\circ\).
We need to find the measure of \(\angle APB\).

Step 2: Key Formula or Approach:
We use the following circle and tangent properties:
1. The radius of a circle is perpendicular to the tangent at the point of contact. Therefore, \(\angle OAP = \angle OBP = 90^\circ\).
2. Radii of the same circle are equal, meaning \(OA = OB\), making triangle \(OAB\) an isosceles triangle.
3. In quadrilateral \(OAPB\), the sum of all interior angles is \(360^\circ\), which implies \(\angle AOB + \angle APB = 180^\circ\) because the other two angles are \(90^\circ\) each.

Step 3: Detailed Explanation:
1. Consider triangle \(OAB\). Since \(OA\) and \(OB\) are both radii of the circle, we have \(OA = OB\).
2. In an isosceles triangle, angles opposite to equal sides are equal. Therefore, \(\angle OBA = \angle OAB = 15^\circ\).
3. By the angle sum property of triangle \(OAB\):
\[ \angle AOB + \angle OAB + \angle OBA = 180^\circ \]
\[ \angle AOB + 15^\circ + 15^\circ = 180^\circ \]
\[ \angle AOB + 30^\circ = 180^\circ \]
\[ \angle AOB = 150^\circ \]
4. We know that the angle between the two tangents drawn from an external point and the angle subtended by the line segments joining the points of contact at the center are supplementary.
\[ \angle APB + \angle AOB = 180^\circ \]
5. Substituting the value of \(\angle AOB\):
\[ \angle APB + 150^\circ = 180^\circ \]
\[ \angle APB = 180^\circ - 150^\circ = 30^\circ \]
6. Thus, \(\angle APB\) is equal to \(30^\circ\).

Step 4: Final Answer:
The correct option is (A).