Question

In a circular museum hall of radius 14 m, some statues are displayed. Statues are kept inside the inner concentric circle of radius 7 m. One such statue lying in sector OAB, is fenced along line segments OA, AP, PB and BO where P is a point on outer circle. Based on above information, answer the following questions:

36(i) Find \(m\angle AOP\).

Hint:

Whenever the adjacent side is exactly half the length of the hypotenuse in a right-angled triangle, the angle between them is always \(60^\circ\) because \(\cos 60^\circ = \frac{1}{2}\).
This is a standard property of \(30^\circ-60^\circ-90^\circ\) triangles.

Answer 1

Correct Answer: 60

Step 1: Understanding the Question:
We are given two concentric circles with center \(O\).
The outer circle has a radius \(R = 14\text{ m}\) and the inner circle has a radius \(r = 7\text{ m}\).
Line segment \(OA\) is a radius of the inner circle, and \(OP\) is a radius of the outer circle.
\(AP\) is tangent to the inner circle at point \(A\), so \(OA \perp AP\) (the angle \(\angle OAP = 90^\circ\)).
We need to find the measure of angle \(\angle AOP\).

Step 2: Key Formula or Approach: In the right-angled triangle \(\Delta OAP\):
\[ \cos \angle AOP = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OA}{OP} \]

Step 3: Detailed Explanation:

• Identify the given side lengths for \(\Delta OAP\):
- Base (radius of inner circle), \(OA = r = 7\text{ m}\)
- Hypotenuse (radius of outer circle), \(OP = R = 14\text{ m}\)

• Since \(AP\) is tangent to the inner circle at \(A\), \(\angle OAP = 90^\circ\).
- Apply trigonometric ratio in right-angled triangle \(\Delta OAP\):
\[ \cos \angle AOP = \frac{OA}{OP} = \frac{7}{14} = \frac{1}{2} \]

• For an acute angle, \(\cos \theta = \frac{1}{2}\) when \(\theta = 60^\circ\).
- Therefore:
\[ m\angle AOP = 60^\circ \]


Step 4: Final Answer:
The measure of \(\angle AOP\) is \(60^\circ\).