In a cyclic quadrilateral `A B C D` , `/_A=(2x+4)o,\ \ /_B=(y+3)o,\ \ /_C=(2y+10)o,\ \ /_D=(4x-5)o` . Find the four angles.

Answer

To solve the problem, we need to find the angles of the cyclic quadrilateral ABCD given the expressions for each angle in terms of variables \(x\) and \(y\). The angles are defined as follows:

- Angle A: \( \angle A = (2x + 4)^\circ \)
- Angle B: \( \angle B = (y + 3)^\circ \)
- Angle C: \( \angle C = (2y + 10)^\circ \)
- Angle D: \( \angle D = (4x - 5)^\circ \)

### Step 1: Use the property of cyclic quadrilaterals
In a cyclic quadrilateral, the sum of the opposite angles is \(180^\circ\). Therefore, we can set up the following equations:

1. \( \angle A + \angle C = 180^\circ \)
2. \( \angle B + \angle D = 180^\circ \)

### Step 2: Substitute the expressions for angles
Substituting the expressions for angles A and C into the first equation:

\[ (2x + 4) + (2y + 10) = 180 \]

Simplifying this gives:

\[ 2x + 2y + 14 = 180 \]

Subtracting 14 from both sides:

\[ 2x + 2y = 166 \]

Dividing the entire equation by 2:

\[ x + y = 83 \quad \text{(Equation 1)} \]

Now, substituting the expressions for angles B and D into the second equation:

\[ (y + 3) + (4x - 5) = 180 \]

Simplifying this gives:

\[ y + 4x - 2 = 180 \]

Adding 2 to both sides:

\[ y + 4x = 182 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:

1. \( x + y = 83 \)
2. \( y + 4x = 182 \)

From Equation 1, we can express \(y\) in terms of \(x\):

\[ y = 83 - x \]

Substituting \(y\) into Equation 2:

\[ (83 - x) + 4x = 182 \]

Combining like terms:

\[ 83 + 3x = 182 \]

Subtracting 83 from both sides:

\[ 3x = 99 \]

Dividing by 3:

\[ x = 33 \]

### Step 4: Find \(y\)
Now substituting \(x = 33\) back into Equation 1 to find \(y\):

\[ 33 + y = 83 \]

Subtracting 33 from both sides:

\[ y = 50 \]

### Step 5: Calculate the angles
Now that we have \(x\) and \(y\), we can find the angles:

1. **Angle A**:
\[ \angle A = 2x + 4 = 2(33) + 4 = 66 + 4 = 70^\circ \]

2. **Angle B**:
\[ \angle B = y + 3 = 50 + 3 = 53^\circ \]

3. **Angle C**:
\[ \angle C = 2y + 10 = 2(50) + 10 = 100 + 10 = 110^\circ \]

4. **Angle D**:
\[ \angle D = 4x - 5 = 4(33) - 5 = 132 - 5 = 127^\circ \]

### Final Angles
Thus, the angles of the cyclic quadrilateral ABCD are:

- \( \angle A = 70^\circ \)
- \( \angle B = 53^\circ \)
- \( \angle C = 110^\circ \)
- \( \angle D = 127^\circ \)