Question

In the adjoining figure, O is the centre of the circle. If \(\angle \text{AOB} = 90^\circ\) and \(\angle \text{ABC} = 30^\circ\), then \(\angle \text{CAO}\) is equal to :

• A. \(30^\circ\)
• B. \(45^\circ\)
• C. \(90^\circ\)
• D. \(60^\circ\)

Hint:

A key theorem for circle geometry is: "The angle at the centre is twice the angle at the circumference subtended by the same arc." For arc AC, \(\angle \text{AOC} = 2 \times \angle \text{ABC}\). Once \(\angle \text{AOC}\) is found, use the property that \(\triangle \text{AOC}\) is isosceles (OA=OC=radii) to find \(\angle \text{CAO}\). Here, \(\angle \text{AOC} = 2 \times 30^\circ = 60^\circ\). Then in isosceles \(\triangle \text{AOC}\), \(\angle \text{CAO} = (180^\circ - 60^\circ)/2 = 60^\circ\).

Answer 1

The Correct Option is D

Concept: This problem uses properties of angles in a circle and properties of triangles. Key theorems include: (A) The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the remaining part of the circle. (B) In a triangle, sides opposite to equal angles are equal, and angles opposite to equal sides (radii in this case) are equal (isosceles triangle property). (C) The sum of angles in a triangle is \(180^\circ\). Step 1: Find \(\angle \text{AOC}\) using \(\angle \text{ABC}\) The arc AC subtends \(\angle \text{AOC}\) at the centre O and \(\angle \text{ABC}\) at point B on the circumference. According to the theorem, the angle at the centre is double the angle at the circumference. \[ \angle \text{AOC} = 2 \times \angle \text{ABC} \] Given \(\angle \text{ABC} = 30^\circ\). \[ \angle \text{AOC} = 2 \times 30^\circ = 60^\circ \] Step 2: Consider triangle AOC In \(\triangle \text{AOC}\): OA and OC are radii of the same circle. Therefore, OA = OC. This means that \(\triangle \text{AOC}\) is an isosceles triangle. Step 3: Find \(\angle \text{CAO}\) in \(\triangle \text{AOC}\) In an isosceles triangle, the angles opposite the equal sides are equal. So, \(\angle \text{OAC} = \angle \text{OCA}\). Let \(\angle \text{OAC} = x\). Then \(\angle \text{OCA} = x\). The angle \(\angle \text{CAO}\) is the same as \(\angle \text{OAC}\). The sum of angles in \(\triangle \text{AOC}\) is \(180^\circ\). \[ \angle \text{OAC} + \angle \text{OCA} + \angle \text{AOC} = 180^\circ \] Substitute the known values: \[ x + x + 60^\circ = 180^\circ \] \[ 2x + 60^\circ = 180^\circ \] \[ 2x = 180^\circ - 60^\circ \] \[ 2x = 120^\circ \] \[ x = \frac{120^\circ}{2} = 60^\circ \] Thus, \(\angle \text{CAO} = 60^\circ\). Step 4: Consistency check with the given \(\angle \text{AOB} = 90^\circ\) This step verifies that the given information \(\angle \text{AOB} = 90^\circ\) is consistent with our result.
In \(\triangle \text{AOB}\), OA = OB (radii), so it's an isosceles triangle. With \(\angle \text{AOB} = 90^\circ\), we have \(\angle \text{OAB} = \angle \text{OBA} = (180^\circ - 90^\circ)/2 = 45^\circ\).
The arc AB subtends \(\angle \text{AOB} = 90^\circ\) at the centre, so it subtends \(\angle \text{ACB} = 90^\circ/2 = 45^\circ\) at the circumference.
In \(\triangle \text{ABC}\), we have \(\angle \text{ACB} = 45^\circ\) and \(\angle \text{ABC} = 30^\circ\) (given). Therefore, \(\angle \text{BAC} = 180^\circ - (45^\circ + 30^\circ) = 180^\circ - 75^\circ = 105^\circ\).
From the figure, \(\angle \text{BAC}\) appears to be \(\angle \text{CAO} + \angle \text{OAB}\). Using our calculated \(\angle \text{CAO} = 60^\circ\) and \(\angle \text{OAB} = 45^\circ\): \(\angle \text{CAO} + \angle \text{OAB} = 60^\circ + 45^\circ = 105^\circ\). This matches the \(\angle \text{BAC}\) calculated from \(\triangle \text{ABC}\), so the given information is consistent. The value of \(\angle \text{CAO}\) is \(60^\circ\). This matches option (4).