Rectangle ABCD with diagonals intersecting at O. If ∠BOC = 44°, find ∠OAD.

Introduction / Context:
Angles formed by diagonals in a rectangle are not fixed at 90° unless it is a square. There is, however, a trigonometric relation between the angle between diagonals and the angle a diagonal makes with a side.

Given Data / Assumptions:

• ABCD is a rectangle; diagonals AC and BD intersect at O.
• ∠BOC = 44°.

Concept / Approach:
Let sides be w (horizontal) and h (vertical). One can show that the angle between diagonals θ satisfies cos θ = (w² − h²)/(w² + h²), and that the angle φ between OA and AD satisfies φ = arctan(w/h). A standard identity yields φ = 90° − θ/2.

Step-by-Step Solution:
Given θ = ∠BOC = 44°.Then ∠OAD = 90° − θ/2 = 90° − 22° = 68°.

Verification / Alternative check:
For a square (θ = 90°), the formula gives ∠OAD = 45°, as expected.

Why Other Options Are Wrong:
90° holds only for a square’s diagonal-to-side angle; 60°/100° conflict with the derived relation.

Common Pitfalls:
Assuming diagonals are perpendicular; only rhombi have perpendicular diagonals, not general rectangles.

Final Answer:
68°