1. **Problem Statement:** In a rectangle ABCD, the diagonals AC and BD intersect at O. Given that $\angle BOC = 40^\circ$, find $\angle OBC$ and $\angle OCB$. 2. **Important Properties:** - In a rectangle, diagonals are equal in length and bisect each other. - The diagonals intersect at point O, so O is the midpoint of both AC and BD. - Triangles formed by the diagonals are isosceles because AO = OC and BO = OD. 3. **Step 1: Understand the triangle BOC** - Since O is the midpoint of BD and AC, BO = OD and AO = OC. - Triangle BOC is isosceles with BO = OC. 4. **Step 2: Use the given angle** - $\angle BOC = 40^\circ$ is the angle between the diagonals at point O. 5. **Step 3: Find the base angles of triangle BOC** - Sum of angles in triangle BOC is $180^\circ$. - Let $\angle OBC = \angle OCB = x$ (since BO = OC, base angles are equal). - So, $x + x + 40^\circ = 180^\circ$. 6. **Step 4: Solve for $x$** $$ 2x + 40 = 180 $$ $$ 2x = 180 - 40 = 140 $$ $$ x = \frac{140}{2} = 70 $$ 7. **Answer:** - $\angle OBC = 70^\circ$ - $\angle OCB = 70^\circ$ Thus, both angles are $70^\circ$ each.