In `Delta XYZ`, I is incentre , O is orthocentre & C is circumcentre . `angle XOZ=2.5 angle XCZ`. Find `angle XIZ`.

Answer

To solve the problem step by step, we will use the properties of triangles and the relationships between the angles given in the question.

### Step 1: Understand the given information
We have triangle XYZ with:
- I as the incenter
- O as the orthocenter
- C as the circumcenter
We are given that:
\[ \angle XOZ = 2.5 \times \angle XCZ \]

### Step 2: Relate angles using circumcenter properties
From the properties of the circumcenter, we know:
\[ \angle XCZ = 2 \times \angle Y \]
This is because the angle at the circumcenter is twice the angle at the opposite vertex.

### Step 3: Relate angles using orthocenter properties
From the properties of the orthocenter, we have:
\[ \angle XOZ = 180^\circ - \angle Y \]
This is because the angle at the orthocenter is supplementary to the angle at the opposite vertex.

### Step 4: Set up the equation
Now we can substitute the expressions for \(\angle XOZ\) and \(\angle XCZ\) into the given equation:
\[ 180^\circ - \angle Y = 2.5 \times (2 \times \angle Y) \]
This simplifies to:
\[ 180^\circ - \angle Y = 5 \times \angle Y \]

### Step 5: Solve for \(\angle Y\)
Rearranging the equation gives us:
\[ 180^\circ = 5 \times \angle Y + \angle Y \]
\[ 180^\circ = 6 \times \angle Y \]
\[ \angle Y = \frac{180^\circ}{6} = 30^\circ \]

### Step 6: Find \(\angle XIZ\)
To find \(\angle XIZ\), we use the formula:
\[ \angle XIZ = 90^\circ + \frac{\angle Y}{2} \]
Substituting the value of \(\angle Y\):
\[ \angle XIZ = 90^\circ + \frac{30^\circ}{2} \]
\[ \angle XIZ = 90^\circ + 15^\circ = 105^\circ \]

### Final Answer
Thus, the value of \(\angle XIZ\) is:
\[ \angle XIZ = 105^\circ \]

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