1. Problem 15 asks: In rectangle ABCD, diagonals AC and BD intersect at O. Given \(\angle CDB = 54^\circ\), find \(\angle OAB\). 2. Important properties: - In a rectangle, diagonals are equal and bisect each other. - Each angle in a rectangle is \(90^\circ\). - Triangles formed by diagonals are congruent. 3. Since \(\angle CDB = 54^\circ\), consider triangle CDB. - Diagonal BD is a straight line from B to D. - \(\angle CDB\) is at vertex D between points C and B. 4. Because ABCD is a rectangle, \(\angle ADC = 90^\circ\). - Triangle CDB is right angled at D. 5. Diagonals bisect each other at O, so \(O\) is midpoint of AC and BD. - \(\angle OAB\) is half of \(\angle CDB\) because of symmetry and congruent triangles formed by diagonals. 6. Therefore, \(\angle OAB = \frac{1}{2} \times 54^\circ = 27^\circ\). Final answer: \(\boxed{27^\circ}\) (Option A).