The diagonals AC and BD of a parallelogram intersects at O. If `angleOAD=40^(@),angleOAB=20^(@)andangleCOD=75^(@)`, then evaluate the following,
`angleACB`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given angles
We have the following angles given:
- \( \angle OAD = 40^\circ \)
- \( \angle OAB = 20^\circ \)
- \( \angle COD = 75^\circ \)

### Step 2: Find \( \angle DAC \)
Since \( O \) is the point of intersection of the diagonals \( AC \) and \( BD \), we can find \( \angle DAC \) using the angles around point \( O \).

We know that:
\[ \angle OAD + \angle OAB + \angle AOB + \angle ODA = 360^\circ \]
However, we can also find \( \angle AOB \) directly from the angles around point \( O \):
\[ \angle AOB = \angle OAD + \angle OAB = 40^\circ + 20^\circ = 60^\circ \]

### Step 3: Find \( \angle AOC \)
The angle \( \angle AOC \) can be found using the property of opposite angles formed by intersecting lines:
\[ \angle AOC = 180^\circ - \angle AOB = 180^\circ - 60^\circ = 120^\circ \]

### Step 4: Find \( \angle COD \)
We already have \( \angle COD = 75^\circ \).

### Step 5: Find \( \angle ACB \)
Using the property of alternate angles in a parallelogram, we know that:
\[ \angle DAC = \angle BCA \]
Since \( \angle OAD = 40^\circ \), we have:
\[ \angle ACB = \angle DAC = 40^\circ \]

### Final Answer
Thus, the value of \( \angle ACB \) is:
\[ \angle ACB = 40^\circ \]

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