Question

A chord \(QR\) subtends an angle of \(105^\circ\) at the centre \(O\) of the circle. The measure of \(\angle RQP\) is

• A. \(\frac{75^\circ}{2}\)
• B. \(\frac{105^\circ}{2}\)
• C. \(75^\circ\)
• D. \(15^\circ\)

Hint:

By the Alternate Segment Theorem, the angle between the tangent \(QP\) and chord \(QR\) is equal to the inscribed angle subtended by the same chord \(QR\) on the major arc.
The inscribed angle is always half of the central angle subtended by the same arc: \[ \angle \text{inscribed} = \frac{\angle \text{centre}}{2} = \frac{105^\circ}{2} \] This provides the answer in just one step!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question describes a circle with centre \(O\). A chord \(QR\) subtends an angle of \(105^\circ\) at the centre \(O\). A tangent line \(QP\) is drawn at the point \(Q\). We need to determine the measure of the angle \(\angle RQP\).

Step 2: Key Formula or Approach:
We can solve this using the Alternate Segment Theorem or basic angle properties of circles: 1. Alternate Segment Theorem: The angle between a chord and a tangent through one of its endpoints is equal to the angle subtended by the chord in the alternate segment.
2. Alternatively, using the radius-tangent relationship: The radius \(OQ\) is perpendicular to the tangent \(QP\) at the point of contact, so: \[ \angle OQP = 90^\circ \] In isosceles triangle \(\triangle OQR\) (where \(OQ = OR = \text{radius}\)), we can find the base angle \(\angle OQR\). Then: \[ \angle RQP = \angle OQP - \angle OQR \]

Step 3: Detailed Explanation:
1. Consider the triangle \(\triangle OQR\): Since \(OQ\) and \(OR\) are both radii of the circle: \[ OQ = OR \] Therefore, \(\triangle OQR\) is an isosceles triangle, which means: \[ \angle OQR = \angle ORQ \] 2. The sum of angles in \(\triangle OQR\) is \(180^\circ\): \[ \angle QOR + \angle OQR + \angle ORQ = 180^\circ \] Substitute the given value \(\angle QOR = 105^\circ\) and use \(\angle OQR = \angle ORQ\): \[ 105^\circ + 2\angle OQR = 180^\circ \] \[ 2\angle OQR = 180^\circ - 105^\circ \] \[ 2\angle OQR = 75^\circ \implies \angle OQR = 37.5^\circ \] 3. Since \(QP\) is a tangent to the circle at point \(Q\), the radius \(OQ\) is perpendicular to \(QP\): \[ \angle OQP = 90^\circ \] 4. Find the measure of \(\angle RQP\): \[ \angle RQP = \angle OQP - \angle OQR \] \[ \angle RQP = 90^\circ - 37.5^\circ = 52.5^\circ \] 5. Writing \(52.5^\circ\) as a fraction: \[ 52.5^\circ = \frac{105^\circ}{2} \]

Step 4: Final Answer:
The measure of \(\angle RQP\) is \(\frac{105^\circ}{2}\).
Thus, the correct option is (B).